Topological Dimension and Dynamical Systems

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Universitext

Michel Coornaert

Topological Dimension and Dynamical Systems

Universitext

Universitext Series editors Sheldon Axler San Francisco State University, San Francisco, CA, USA Vincenzo Capasso Università degli Studi di Milano, Milan, Italy Carles Casacuberta Universitat de Barcelona, Barcelona, Spain Angus MacIntyre Queen Mary University of London, London, UK Kenneth Ribet University of California, Berkeley, CA, USA Claude Sabbah CNRS École Polytechnique Centre de mathématiques, Palaiseau, France Endre Süli University of Oxford, Oxford, UK Wojbor A. Woyczyński Case Western Reserve University, Cleveland, OH, USA

Universitext is a series of textbooks that presents material from a wide variety of mathematical disciplines at master’s level and beyond. The books, often well class-tested by their author, may have an informal, personal, even experimental approach to their subject matter. Some of the most successful and established books in the series have evolved through several editions, always following the evolution of teaching curricula, into very polished texts. Thus as research topics trickle down into graduate-level teaching, ﬁrst textbooks written for new, cutting-edge courses may make their way into Universitext.

More information about this series at http://www.springer.com/series/223

Michel Coornaert

Topological Dimension and Dynamical Systems

123

Michel Coornaert Institut de Recherche Mathématique Avancée University of Strasbourg Strasbourg France

ISSN 0172-5939 Universitext ISBN 978-3-319-19793-7 DOI 10.1007/978-3-319-19794-4

ISSN 2191-6675

(electronic)

ISBN 978-3-319-19794-4

(eBook)

Library of Congress Control Number: 2015941878 Mathematics Subject Classiﬁcation: 54F45, 37B05, 37B10, 37B40, 54H20, 43A07 Springer Cham Heidelberg New York Dordrecht London © Springer International Publishing Switzerland 2015 Translation from the French language edition: Dimension Topologique et Systèmes Dynamiques by Michel Coornaert, © Société Mathématique de France 2005. All rights reserved This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper Springer International Publishing AG Switzerland is part of Springer Science+Business Media (www.springer.com)

To Marianne

Preface to the English Edition

This is a revised and augmented English edition of my book “Dimension topologique et systèmes dynamiques” which was published in 2005 by the Société Mathématique de France. As explained in the preface to the French edition, the goal of the book is to provide a self-contained introduction to mean topological dimension, an invariant of dynamical systems introduced in 1999 by Misha Gromov, and explain how this invariant was successfully used by Elon Lindenstrauss and Benjamin Weiss to answer a long-standing open question about embeddings of minimal dynamical systems into shifts. A large number of revisions and additions have been made to the original text. Chapter 5 contains an entirely new section devoted to the Sorgenfrey line. Two chapters have also been added: Chap. 9 on amenable groups and Chap. 10 on mean topological dimension for continuous actions of countable amenable groups. These new chapters contain material that has never before appeared in textbook form. The chapter on amenable groups is based on Følner’s characterization of amenability and may be read independently from the rest of the book. There are a total of 160 exercises. The hardest ones are accompanied with hints. Although the contents of this book lead directly to several active areas of current research in mathematics and mathematical physics, the prerequisites needed for reading it remain modest, essentially some familiarities with undergraduate point-set topology and, in order to access the ﬁnal two chapters, some acquaintance with basic notions in group theory. There are many people I would like to thank for their assistance during the preparation of this book: Insa Badji, Nathalie Coornaert and Lindzy Tossé for helping me in drawing the ﬁgures; Fabrice Krieger and Tullio Ceccherini-Silberstein for proofreading the manuscript and offering invaluable suggestions; Dr. Jeorg Sixt, Catherine Waite, and the editorial staff at Springer-Verlag for their competence and guidance during the publication process. Finally, I want to thank my wife Martine for her patience and understanding while this book was being written. Strasbourg October 2014

Michel Coornaert

vii

Preface to the French Edition

This book grew out from a DEA course I gave at the University of Strasbourg in Spring 2002. The ﬁrst part of the book presents some fundamental results from dimension theory. The second part is devoted to topological mean dimension and its applications to embeddings problems for dynamical systems. Dimension theory is the branch of general topology that studies the notion of dimension for topological spaces. It has its root at the origins of geometry and the difﬁculties encountered by mathematicians when trying to give rigorous deﬁnitions of the concepts of curves and surfaces. The theory flourished at the end of the nineteenth century and at the beginning of the twentieth century. Its developments had a deep impact on many other branches of mathematics such as algebraic topology, dynamical systems, and probability theory. Actually, several non-equivalent deﬁnitions of dimension for topological spaces may be found in the literature. The most commonly used are the small inductive dimension ind, the large inductive dimension Ind, and the covering dimension dim. Small inductive dimension was introduced by P. Urysohn in 1922 and independently by K. Menger in 1923. Large inductive dimension and covering dimension were introduced by E. Čech in 1931. These three dimensions coincide for separable metrizable spaces. Mean topological dimension is a conjugacy invariant of topological dynamical systems which was recently introduced by Gromov [44]. This invariant enables one to distinguish systems with inﬁnite topological entropy. It was used by Lindenstrauss and Weiss [74] to answer a long-standing open question about the existence of embeddings of minimal dynamical systems into shifts. Chapter 1 begins with the deﬁnition of the covering dimension of a topological space and the proof of its main properties. We establish in particular the countable union theorem in normal spaces and the monotonicity theorem in metric spaces. The second chapter is devoted to 0-dimensional topological spaces. Examples of such spaces are given and we investigate the relationship between the class of 0-dimensional spaces and other classes of highly disconnected topological spaces. The notion of a polyhedron is introduced in Chap. 3. A polyhedron is a topological space that can be triangulated, i.e., is homeomorphic to the geometric realization of some ﬁnite simplicial complex. We prove Lebesgue’s lemma on open ix

x

Preface to the French Edition

covers of Euclidean cubes. It is used to show that the covering dimension of a polyhedron is equal to the combinatorial dimension of any of its triangulations. We also deduce from Lebesgue’s lemma that the covering dimension of Rn is equal to n as expected. In Chap. 4, we prove Aleksandrov theorem about topological dimension of compact metrizable spaces and ε-injective maps. We then establish the Menger-Nöbeling embedding theorem that states that any n-dimensional compact metrizable space can be embedded in R2nþ1 . Chapter 5 is devoted to the study of counterexamples which played an important role in the history of dimension theory: Erdös and Bing spaces, Knaster-Kuratowski fan, Tychonoff plank. These counterexamples enlighten the validity domains of some of the theorems established in the previous chapters. In Chap. 6, the mean topological dimension mdimðX; TÞ of a discrete dynamical system ðX; TÞ, where X is a normal space and T : X ! X a continuous map, is deﬁned and its ﬁrst properties are established. When X is a compact metric space, an equivalent deﬁnition of mdimðX; TÞ involving the metric is given. In Chap. 7, we consider the dynamical system ðK Z ; σÞ, where K Z is the space of bi-inﬁnite sequences of points in a topological space K and σ is the shift map ðxi Þ 7! ðxiþ1 Þ. We show that mdimðK Z ; σÞ dimðKÞ for any compact metrizable space K and that equality holds when K is a polyhedron. By considering appropriate subshifts, we show that mean topological dimension can take any value in ½0; 1. Chapter 8 discusses embeddings problems of dynamical systems into shifts. We prove the theorem of Jaworski that asserts that any dynamical system ðX; TÞ, where T is a homeomorphism without periodic points of a ﬁnite-dimensional compact metrizable space X, can be embedded into the shift ðRZ ; σÞ. Finally, we describe the Lindenstrauss-Weiss counterexamples which show that Jaworski’s theorem becomes false if the hypothesis on the ﬁniteness of the topological dimension is removed. There are historical notes and a list of exercises at the end of each chapter. All along the text, I tried to give detailed proofs in order to make them accessible to students who attended a basic course on general topology. The terminology used is that of Bourbaki with the exception that compact (resp. locally compact, resp. normal, resp. scattered) spaces are not required to be Hausdorff. I thank all the students who attended my course and especially Fabrice Krieger for numerous suggestions. I am also very grateful to Nathalie Coornaert, Lida Leyva, and Stéphane Laurent who helped me in the preparation of the manuscript and the realization of the ﬁgures.

Contents

Part I

Topological Dimension

1

Topological Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Definition of Topological Dimension. . . . . . . . . . . . . . 1.2 Topological Dimension of Closed Subsets . . . . . . . . . . 1.3 Topological Dimension of Connected Spaces . . . . . . . . 1.4 Topological Dimension of Compact Metric Spaces . . . . 1.5 Normal Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Topological Dimension of Normal Spaces . . . . . . . . . . 1.7 The Countable Union Theorem. . . . . . . . . . . . . . . . . . 1.8 Topological Dimension of Subsets of Metrizable Spaces Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Zero-Dimensional Spaces . . . . . . . . . . . . . . . . . . . 2.1 The Cantor Set . . . . . . . . . . . . . . . . . . . . . . . 2.2 Scattered Spaces . . . . . . . . . . . . . . . . . . . . . . 2.3 Scatteredness of Zero-Dimensional Spaces . . . . 2.4 Lindelöf Spaces . . . . . . . . . . . . . . . . . . . . . . 2.5 Totally Disconnected Spaces . . . . . . . . . . . . . 2.6 Totally Separated Spaces . . . . . . . . . . . . . . . . 2.7 Zero-Dimensional Compact Hausdorff Spaces . 2.8 Zero-Dimensional Separable Metrizable Spaces 2.9 Zero-Dimensional Compact Metrizable Spaces . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Topological Dimension of Polyhedra 3.1 Simplices of Rn . . . . . . . . . . . 3.2 Simplicial Complexes of Rn . . . 3.3 Open Stars . . . . . . . . . . . . . . .

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Contents

3.4 Barycentric Subdivisions . . . . . . . . . . . . . 3.5 The Lebesgue Lemma and its Applications 3.6 Abstract Simplicial Complexes . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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54 60 64 65 66

4

Dimension and Maps . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 The Tietze Extension Theorem . . . . . . . . . . . . . . . . 4.2 Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Nerve of a Cover . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 α-Compatible Maps. . . . . . . . . . . . . . . . . . . . . . . . 4.5 "-Injective Maps . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Definition of dimε ðX; dÞ . . . . . . . . . . . . . . . . . . . . 4.7 Euclidean Embeddings of Finite-Dimensional Spaces Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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69 69 72 73 74 76 78 80 83 84

5

Some Classical Counterexamples . 5.1 The Erdös Space. . . . . . . . . 5.2 The Knaster-Kuratowski Fan 5.3 The Bing Space . . . . . . . . . 5.4 The Tychonoff Plank. . . . . . 5.5 The Sorgenfrey Plane . . . . . Notes . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . .

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87 87 90 93 95 99 101 102

6

Mean Topological Dimension for Continuous Maps . . . 6.1 Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Subadditive Sequences . . . . . . . . . . . . . . . . . . . . 6.3 Definition of Mean Topological Dimension . . . . . . 6.4 General Properties of Mean Topological dimension . 6.5 Metric Approach to Mean Topological Dimension . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7

Shifts 7.1 7.2 7.3 7.4

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123 123 125 127 129

Part II

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Mean Topological Dimension

and Subshifts over Z . . . . Shifts . . . . . . . . . . . . . . . Shifts on Polyhedra . . . . . . Mean Projective Dimension Subshifts of Finite Type . .

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Contents

7.5 7.6

Subshifts of Block-Type . . . . Construction of Subshifts with Dimension . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . .

xiii

............. Prescribed Mean ............. ............. .............

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8

Applications of Mean Dimension to Embedding Problems 8.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Embeddings in Shifts . . . . . . . . . . . . . . . . . . . . . . . 8.3 Jaworski’s Embedding Theorem . . . . . . . . . . . . . . . . 8.4 The Lindenstrauss-Weiss Counterexamples. . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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139 139 141 143 145 153 153

9

Amenable Groups . . . . . . . . . . . . . . . . . . . . 9.1 Følner Sequences . . . . . . . . . . . . . . . . 9.2 Amenable Groups . . . . . . . . . . . . . . . . 9.3 Examples of Non-amenable Groups. . . . 9.4 The Subadditive Convergence Theorem for Amenable Groups . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 Mean Topological Dimension for Actions of Amenable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Continuous Actions. . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Definition of Mean Topological Dimension . . . . . . . . 10.3 General Properties of Mean Topological Dimension . . 10.4 Metric Approach to Mean Topological Dimension . . . 10.5 Shifts and Subshifts . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Mean Topological Dimension of Shifts over Polyhedra 10.7 Subshifts of Block-Type . . . . . . . . . . . . . . . . . . . . . 10.8 Construction of Subshifts with Prescribed Mean Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

223

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

229

Symbols

#E ordx ðαÞ ordðαÞ βα DðαÞ dimðXÞ indðXÞ IndðXÞ cdimG ðXÞ jC j StC ðpÞ kf kX1 suppðf Þ α_β stabdimðXÞ mdimðX; TÞ σ ¼ σK prodimðXÞ jAj P fin ðGÞ AΔB mdimðX; G; TÞ Σ

Cardinality of E Order of α at x Order of α β ﬁner than α Minimum of ordðβÞ for β α Covering dimension of X Small inductive dimension of X Large inductive dimension of X Cohomological dimension of X with coefﬁcients in G Support of the simplicial complex C Open star of C at p Sup-norm of f Support of f Join of α and β Stable dimension of X Mean topological dimension The shift map on K Z Mean projective dimension Cardinality of A Set of all ﬁnite subsets of G Symmetric difference of A and B Mean topological dimension Shift action on K G

xv

Part I

Topological Dimension

Chapter 1

Topological Dimension

In this chapter, we introduce the topological dimension dim(X ), also called the covering dimension, of a topological space X . The definition of dim(X ) involves the combinatorics of the finite open covers of X . We establish some basic properties of the topological dimension and give first examples of topological spaces for which it can be explicitly computed. We introduce the class of normal spaces and prove the countable union theorem for closed subsets of normal spaces (Theorem 1.7.1). We also show the monotonicity of topological dimension for subsets of metrizable spaces (Theorem 1.8.3).

1.1 Definition of Topological Dimension We denote by N the set of non-negative integers and by R the set of real numbers. We use the symbol ∞ with the usual conventions ∞ + ∞ = ∞, ∞ + a = a + ∞ = ∞ and a < ∞ for all a ∈ R. When E is a set, #E denotes the cardinality of E if E is finite or the symbol ∞ otherwise. Let X be a set. Let α = (Ai )i∈I be a family of subsets of X indexed by a set I . For each x ∈ X , let ord x (α) := −1 + #{i ∈ I | x ∈ Ai }. We say that the quantity ord x (α) ∈ {−1} ∪ N ∪ {∞} is the order of α at the point x. We define the (global) order ord(α) ∈ {−1} ∪ N ∪ {∞} of the family α by ord(α) := sup ord x (α). x∈X

(If X is the empty set ∅, we adopt the convention ord(α) = −1.) In other words, ord(α) is the greatest integer n (or ∞ if such an integer does not exist) such that we can find n + 1 distinct elements i 0 , i 1 , . . . , i n ∈ I satisfying © Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_1

3

4

1 Topological Dimension

Ai0 ∩ Ai1 ∩ · · · ∩ Ain = ∅. We say that α is a cover of X (or that α covers X ) if we have X=

Ai .

i∈I

Note that α is a cover of X if and only if one has ord x (α) ≥ 0 for all x ∈ X . We say that α is a partition of X if α is a cover of X and Ai ∩ A j = ∅ for all distinct i, j ∈ I . Note that the family α is a partition of X if and only if one has ord x (α) = 0 for all x ∈ X . Let α = (Ai )i∈I and β = (B j ) j∈J be two covers of a set X . We say that the cover β is finer than the cover α, and we write β α, if for every j ∈ J there exists i ∈ I such that B j ⊂ Ai . If α, β and γ are covers of a set X such that γ β and β α, then we clearly have γ α (transitivity of ). Remark 1.1.1 Let α = (Ai )i∈I and β = (Bi )i∈I be covers of a set X such that Bi ⊂ Ai for all i ∈ I . Then β is finer than α. Moreover, for each x ∈ X , we have {i ∈ I | x ∈ Bi } ⊂ {i ∈ I | x ∈ Ai } and hence ord x (β) ≤ ord x (α). Consequently, we have ord(β) ≤ ord(α). We say that a cover α = (Ai )i∈I of a topological space X is an open cover (resp. a closed cover) of X if Ai is open (resp. closed) in X for all i ∈ I . Definition 1.1.2 Let X be a topological space. Let α = (Ui )i∈I be a finite open cover of X . We define the quantity D(α) by D(α) := min ord(β), β

where β runs over all finite open covers of X that are finer than α. Remark 1.1.3 (1) As α α, we have D(α) ≤ ord(α) ≤ −1 + #I . (2) We have D(α) ∈ {−1} ∪ N. (3) We have D(α) ≤ n if and only if there exists a finite open cover β α such that ord(β) ≤ n. Proposition 1.1.4 Let X be a topological space. Let α and α be finite open covers of X such that α α . Then one has D(α) ≥ D(α ). Proof If β is a finite open cover of X such that β α, then we have β α by transitivity of . Consequently, we have D(α) ≥ D(α ). We shall use the following auxiliary result.

1.1 Definition of Topological Dimension

5

Lemma 1.1.5 Let X be a topological space. Let α = (Ui )i∈I be a cover of X and let β = (V j ) j∈J be an open cover of X such that β α. Then there exists an open cover γ = (Wi )i∈I of X such that ord x (γ) ≤ ord x (β) for all x ∈ X and Wi ⊂ Ui for all i ∈ I . Proof As β is finer than α, there exists a map ϕ : J → I such that V j ⊂ Uϕ( j) for all j ∈ J . Consider the family γ = (Wi )i∈I defined by Wi :=

Vj.

j∈ϕ−1 (i)

Each Wi is open in X since it is a union of open subsets of X . On the other hand, we have Wi ⊂ Ui as V j ⊂ Ui for all j ∈ ϕ−1 (i). As β = (V j ) j∈J covers X and V j ⊂ Wϕ( j) for all j ∈ J , we deduce that γ is a cover of X . Finally, consider a point x ∈ X . We have x ∈ Wi if and only if there exists j ∈ ϕ−1 (i) such that x ∈ V j . It follows that ϕ induces a surjection from { j ∈ J | x ∈ V j } onto {i ∈ I | x ∈ Wi }. This implies ord x (γ) ≤ ord x (β). Proposition 1.1.6 Let X be a topological space and let α = (Ui )i∈I be a finite open cover of X . Then one has D(α) = min ord(β), β

where β runs over all open covers (finite or not) of X that are finer than α. Proof It suffices to show that every open cover of X that is finer than α has order at least D(α). Let β be an open cover of X such that β α. By Lemma 1.1.5, there exists an open cover γ = (Wi )i∈I such that ord(γ) ≤ ord(β) and Wi ⊂ Ui for all i ∈ I . As γ is a finite open cover of X that is finer than α, we have D(α) ≤ ord(γ) and hence D(α) ≤ ord(β). Proposition 1.1.7 Let X be a topological space and let α = (Ui )i∈I be a finite open cover of X . Then one has D(α) = min ord(β), β

where β runs over all finite open covers of X that are of the form β = (Vi )i∈I with Vi ⊂ Ui for all i ∈ I . Proof This is again an immediate consequence of Lemma 1.1.5. Indeed, this lemma implies that, for every finite open cover γ of X that is finer than α, there exists an open cover β = (Vi )i∈I such that ord(β) ≤ ord(γ) and Vi ⊂ Ui for all i ∈ I . Definition 1.1.8 Let X be a topological space. The topological dimension dim(X ) ∈ {−1} ∪ N ∪ {∞} of X is the quantity defined by dim(X ) := sup D(α), α

where α runs over all finite open covers of X .

6

1 Topological Dimension

ˇ The topological dimension dim(X ) is also called the Cech-Lebesgue covering dimension, or simply the Lebesgue dimension, of X . It is clear from its definition that topological dimension is a topological invariant, that is, one has dim(X ) = dim(Y ) whenever X and Y are homeomorphic topological spaces. Example 1.1.9 One has dim(X ) = −1 if and only if X = ∅. Example 1.1.10 Let X be a non-empty set equipped with its discrete topology, i.e., with the topology for which all subsets of X are open. Let α be a finite open cover of X . Then the family β = {x} x∈X is an open partition of X . As β α, it follows from Proposition 1.1.6 that D(α) = 0. Consequently, we have dim(X ) = 0. Example 1.1.11 Let n ∈ N and let X be a finite set of cardinality n + 2. Fix an arbitrary element x0 ∈ X and equip X with the topology for which the open subsets are ∅ and all the subsets of X containing x0 . Then α := ({x0 , x})x∈X \{x0 } is a finite open cover of X . Observe that any open cover of X that is finer than α must contain {x0 , x} for each x ∈ X \{x0 }. It follows that D(α) = ord(α) = n. As α is finer than any open cover of X , we conclude that dim(X ) = D(α) = n. Example 1.1.12 Let X be an infinite set. Let us equip X with its cofinite topology, i.e., the topology for which the open sets are the empty set and all the subsets U ⊂ X with X \U finite. The space X is not Hausdorff since the intersection of two nonempty open subsets of X is never empty. Note that X is compact. Indeed, if γ is an open cover of X , a finite subcover of γ may be obtained by choosing a non-empty open subset U of γ and then, for each x ∈ X \U , an open subset of γ containing x. Let us show that dim(X ) = ∞. Let n ∈ N and let F be a subset of X with cardinality n + 1. Consider the finite open cover α = (Ux )x∈F , where Ux := (X \F) ∪ {x} = X \(F\{x}) for all x ∈ F. Suppose that β = (V j ) j∈J is a finite open cover of X that is finer than α. Then any element of β contains at most one point belonging to F. Therefore, we can find n + 1 distinct elements j0 , j1 , . . . , jn ∈ J such that the open sets V j0 , V j1 , . . . , V jn are all non-empty. As a finite intersection of non-empty open subsets of X is never empty, we deduce that ord(β) ≥ n. Therefore, we have D(α) ≥ n. This implies dim(X ) ≥ n for all n ∈ N, so that dim(X ) = ∞. Proposition 1.1.13 Let X be a topological space. Let n ∈ N. Then the following conditions are equivalent: (a) dim(X ) ≤ n; (b) for every finite open cover α of X , there exists a finite open cover β α such that ord(β) ≤ n;

1.1 Definition of Topological Dimension

7

(c) for every finite open cover α of X , there exists an open cover β α such that ord(β) ≤ n; (d) for every finite open cover α = (Ui )i∈I of X , there exists an open cover β = (Vi )i∈I of X such that ord(β) ≤ n and Vi ⊂ Ui for all i ∈ I . Proof This immediately follows from Definitions 1.1.2, 1.1.8, Propositions 1.1.6, and 1.1.7.

1.2 Topological Dimension of Closed Subsets If Y is a subset of a topological space, then Y is itself a topological space for the induced topology, i.e., the topology on Y for which the open sets are all the sets of the form U = V ∩ Y , where V is an open subset of X . It is natural to try to investigate the relations between the topological dimension of a topological space and the topological dimension of its subsets. For closed subsets, we have the following general result. Proposition 1.2.1 Let X be a topological space and let F ⊂ X be a closed subset. Then one has dim(F) ≤ dim(X ). Proof Let α = (Ui )i∈I be a finite open cover of F. By definition of the induced topology, we can find, for each i ∈ I , an open subset Vi of X such that Ui = Vi ∩ F. Then the family β := (Vi )i∈I ∪ {X \F} is a finite open cover of X . Therefore, there exists a finite open cover γ = (W j ) j∈J of X with γ β and ord(γ) ≤ dim(X ). Clearly γ := (W j ∩ F) j∈J is a finite open cover of F that is finer than α. Moreover, we have ord x (γ ) = ord x (γ) for all x ∈ F and hence ord(γ ) ≤ ord(γ) ≤ dim(X ). It follows that D(α) ≤ dim(X ). Consequently, we have dim(F) = supα D(α) ≤ dim(X ). When Y is a subset of a topological space X , it may happen that dim(Y ) > dim(X ) (“non-monotonicity” of the topological dimension). To provide such an example, it suffices to start from a topological space Y with positive topological dimension and then embed it in a zero-dimensional space X by using the following construction. Example 1.2.2 Let Y be a topological space and let X := Y ∪{x0 } be the set obtained / Y . Equip X with the topology for which the from Y by adjoining an element x0 ∈ open subsets are X and all the subsets ⊂ Y that are open with respect to the initial topology on Y . Observe that the topology induced by X on Y is its initial topology. On the other hand, the open cover of X that is reduced to X is finer than any open cover of X since X is the only open subset of X that contains x0 . It follows that dim(X ) = 0. Remark 1.2.3 Observe that the space X in the previous example is never Hausdorff for Y = ∅. In Sect. 5.4, we will give an example of a compact Hausdorff space X with dim(X ) = 0 containing a subset Y ⊂ X with dim(Y ) > 0. However, we will

8

1 Topological Dimension

see at the end of this chapter that every subset Y of a metrizable space X satisfies dim(Y ) ≤ dim(X ) (Theorem 1.8.3). Lemma 1.2.4 Let X be a topological space and F a closed subset of X . Let α = (Ui )i∈I be a finite open cover of X . Then there exists an open cover β = (Vi )i∈I of X such that Vi ⊂ Ui for all i ∈ I and ord x (β) ≤ dim(F) for all x ∈ F. Proof The family γ := (F ∩ Ui )i∈I is a finite open cover of F. Therefore, we have D(γ) ≤ dim(F). By Proposition 1.1.7, we can find an open cover δ = (Wi )i∈I of F such that Wi ⊂ F ∩ Ui for all i ∈ I and ord(δ) = D(γ) ≤ dim(F). The sets Wi are open subsets of F. Thus, for each i ∈ I , there exists an open subset i of X such that Wi = F ∩ i . Consider now the family β = (Vi )i∈I consisting of the subsets of X defined by Vi := i ∪ (X \F) ∩ Ui for all i ∈ I . Clearly β is an open cover of X satisfying Vi ⊂ Ui for all i ∈ I . Moreover, we have F ∩ Vi = F ∩ i = Wi for all i ∈ I . It follows that, for all x ∈ F, ord x (β) = ord x (δ) ≤ ord(δ) ≤ dim(F). This shows that the cover β has the required properties.

Proposition 1.2.5 Let X be a topological space. Let F and G be closed subsets of X such that X = F ∪ G. Then one has dim(X ) = max(dim(F), dim(G)). Proof We have dim(X ) ≥ dim(F) and dim(X ) ≥ dim(G) by Proposition 1.2.1. Thus, it suffices to prove that dim(X ) ≤ max(dim(F), dim(G)). Let α = (Ui )i∈I be a finite open cover of X . By virtue of Lemma 1.2.4, there exists an open cover β = (Vi )i∈I of X such that Vi ⊂ Ui for all i ∈ I and ord x (β) ≤ dim(F) for all x ∈ F. By applying again Lemma 1.2.4, we can find an open cover γ = (Wi )i∈I of X such that Wi ⊂ Vi for all i ∈ I and ord x (γ) ≤ dim(G)

for all x ∈ G.

(1.2.1)

As Wi ⊂ Vi for all i ∈ I , we have ord x (γ) ≤ ord x (β) for all x ∈ X . We deduce that ord x (γ) ≤ ord x (β) ≤ dim(F)

for all x ∈ F.

(1.2.2)

As X = F ∪ G, inequalities (1.2.1) and (1.2.2) imply that ord(γ) ≤ max(dim(F), dim(G)). Now observe that the cover γ is finer than α since Wi ⊂ Vi ⊂ Ui for all i ∈ I . It follows that D(α) ≤ max(dim(F), dim(G)). Consequently, we have dim(X ) = supα D(α) ≤ max(dim(F), dim(G)).

1.2 Topological Dimension of Closed Subsets

9

By an immediate induction on the integer n, we get the following result. Corollary 1.2.6 Let X be a topological space and let F1 , . . . , Fn (n ≥ 1) be closed subsets of X such that X=

Fk .

1≤k≤n

Then one has dim(X ) = max dim(Fk ). 1≤k≤n

1.3 Topological Dimension of Connected Spaces Recall that a topological space X is said to be connected if the only subsets of X that are both open and closed are ∅ and X . Definition 1.3.1 One says that a topological space X is accessible if every subset of X that is reduced to a single point is closed in X . Accessible spaces are also called T1 -spaces. A topological space X is accessible if and only if, given any pair of distinct points x and y in X , there exists a neighborhood of x that does not contains y. Every Hausdorff space is clearly accessible. The converse implication is false as shown by the following example. Example 1.3.2 Take again an infinite set X equipped with its cofinite topology as in Example 1.1.12. The closed subsets of X are X and all its finite subsets. Therefore X is accessible. However, as we have already observed in Example 1.1.12, X is not Hausdorff since any two non-empty open subsets of X always meet. Note that X is connected and compact. Proposition 1.3.3 Let X be a connected accessible topological space containing more than one point. Then one has dim(X ) ≥ 1. Proof Let x and y be two distinct points in X . As X is accessible, the subsets X \{x} and X \{y} are open in X . Consider the open cover α = X \{x}, X \{y} . The connectedness of X implies that every open partition of X is trivial. It follows that D(α) ≥ 1. Since dim(X ) ≥ D(α), we conclude that dim(X ) ≥ 1. Proposition 1.3.3 becomes false if we remove the accessibility hypothesis. Indeed, consider a set X , containing at least two distinct points, equipped with its trivial topology, i.e., the topology for which the only open subsets are ∅ and X . Then X is connected. However, we have dim(X ) = 0 since the trivial open cover{X }, which has order 0, is finer than any finite open cover of X .

10

1 Topological Dimension

1.4 Topological Dimension of Compact Metric Spaces Let (X, d) be a metric space. For x ∈ X and r > 0, we denote by B(x, r ) the open ball of radius r centered at x. The diameter diam(Y ) of a subset Y ⊂ X is diam(Y ) := sup d(y1 , y2 ) ∈ [0, ∞]. y1 ,y2 ∈Y

We define the mesh of a cover α = (Ai )i∈I of X by mesh(α) := sup diam(Ai ) ∈ [0, ∞]. i∈I

Remark 1.4.1 If α and β are covers of a metric space such that β α, then one has mesh(β) ≤ mesh(α). Proposition 1.4.2 Let α = (Ui )i∈I be an open cover of a compact metric space X . Then there exists a real number λ > 0 satisfying the following property: for every subset Y ⊂ X such that diam(Y ) ≤ λ, there exists i ∈ I such that Y ⊂ Ui . Proof Let us choose, for each x ∈ X , an index i(x) ∈ I such that x ∈ Ui(x) . As Ui(x) is an open subset, there exists a real number r x > 0 such that the open ball B(x, 2r x ) is entirely contained in Ui(x) . The open balls B(x, r x ), x ∈ X , cover X . By compactness of X , there exists a finite subset A ⊂ X such that the balls B(x, r x ), x ∈ A, cover X . Let us set λ := min x∈A r x . We have λ > 0. Suppose that Y ⊂ X satisfies diam(Y ) ≤ λ and choose an arbitrary point y ∈ Y . Then we can find a point a ∈ A such that d(a, y) < ra . By applying the triangle inequality, we get Y ⊂ B(a, ra + λ) ⊂ B(a, 2ra ) ⊂ Ui(a) . Consequently, λ has the required property.

A real number λ > 0 satisfying the condition of Proposition 1.4.2 is called a Lebesgue number of the open cover α. Corollary 1.4.3 Let α be an open cover of a compact metric space X . Then there exists a real number λ > 0 such that every cover β of X with mesh(β) ≤ λ satisfies β α. Proof We can take as λ any Lebesgue number of α.

Proposition 1.4.4 Let X be a compact metric space and let n ∈ N. Then the following conditions are equivalent: (a) dim(X ) ≤ n; (b) for every ε > 0, there exists a finite open cover α of X such that mesh(α) ≤ ε and ord(α) ≤ n;

1.4 Topological Dimension of Compact Metric Spaces

11

Fig. 1.1 The open cover α3 = {[x0 , x2 ), (x1 , x3 ), (x2 , x4 ), (x3 , x5 ), (x4 , x6 ]}

(c) there exists a sequence (αk )k∈N of finite open covers of X such that limk→∞ mesh(αk ) = 0 and ord(αk ) ≤ n for all k ∈ N. Proof Conditions (b) and (c) are clearly equivalent. Let ε > 0. As X is compact, there exists a finite cover α of X consisting of open balls of radius ε/2. If dim(X ) ≤ n, we can find a finite open cover β α with ord(β) ≤ n. We then have mesh(β) ≤ mesh(α) ≤ ε. This shows that (a) implies (b). Conversely, suppose (b). Let α be a finite open cover of X . Let λ > 0 be a Lebesgue number of α. Since condition (b) is satisfied, there exists a finite open cover β of X such that mesh(β) ≤ λ and ord(β) ≤ n. This implies β α and hence D(α) ≤ n. We deduce that dim(X ) ≤ n. This shows that (b) implies (a). Let us use the above results to determine the topological dimension of the unit segment [0, 1] ⊂ R. Note that any segment of R, and, more generally, any segment of a Hausdorff topological vector space, is homeomorphic to [0, 1] and hence has the same dimension. Proposition 1.4.5 The unit segment [0, 1] ⊂ R has topological dimension dim ([0, 1]) = 1. Proof As [0, 1] is connected, we have dim([0, 1]) ≥ 1 by Proposition 1.3.3. Let k ≥ 2 be an integer. Consider the points xi ∈ [0, 1] defined by xi =

i for all i ∈ {0, 1, . . . , 2k}. 2k

Let αk be the finite open cover of [0, 1] consisting of the intervals [x0 , x2 ), (x2k−2 , x2k ] and all the intervals of the form (xi , xi+2 ) for i ∈ {1, . . . , 2k − 3} (see Fig. 1.1 for k = 3). We have that ord(αk ) = 1. On the other hand, mesh(αk ) = 1/k tends to 0 as k goes to infinity. This implies dim([0, 1]) ≤ 1 by Proposition 1.4.4.

1.5 Normal Spaces In this section, we introduce the important class of normal topological spaces. We shall see in particular that all metrizable spaces and all compact Hausdorff spaces are normal.

12

1 Topological Dimension

Fig. 1.2 Separation of closed subsets in a normal space

Definition 1.5.1 One says that a topological space X is normal if, given any pair of disjoint closed subsets A and B of X , there exist disjoint open subsets U and V of X such that A ⊂ U and B ⊂ V (Fig. 1.2). Note that every accessible normal space is Hausdorff. Normal Hausdorff spaces are also called T4 -spaces. Proposition 1.5.2 Let X be a topological space. Then the following conditions are equivalent: (a) X is normal; (b) for every closed subset A ⊂ X and every open subset U ⊂ X such that A ⊂ U , there exists an open subset V ⊂ X such that A ⊂ V ⊂ V ⊂ U (here V denotes the closure of V in X ). Proof Let us first show that (a) implies (b). Let X be a normal space. Let A be a closed subset and U an open subset with A ⊂ U . Then the set B = X \U is closed in X and does not meet A. Therefore we can find disjoint subsets V and W of X such that A ⊂ V and B ⊂ W . The set V is contained in the closed subset X \W . It follows that V ⊂ X \W ⊂ U . This shows that X satisfies (b). Conversely, suppose that X satisfies (b). Let A and B be disjoint closed subsets of X . Then the open subset U := X \B satisfies A ⊂ U . By (b), there exists an open subset V ⊂ X such that A ⊂ V ⊂ V ⊂ U . Observe now that the open subsets V and X \V are disjoint and contain A and B respectively. This shows that X is normal. Proposition 1.5.3 Every metrizable space is normal. Proof Let (X, d) be a metric space. Let Y be a non-empty subset of X . The distance of a point x ∈ X to Y is dist(x, Y ) := inf d(x, y). y∈Y

It follows from the triangle inequality that the map x → dist(x, Y ) is 1-Lipschitz and hence continuous on X . Moreover, one has dist(x, Y ) = 0 if and only if x ∈ Y .

1.5 Normal Spaces

13

Suppose now that A and B are disjoint non-empty closed subsets of X . The map f : X → R defined by f (x) := dist(x, A) − dist(x, B) is continuous. The open subsets U := {x ∈ X | f (x) < 0} and V := {x ∈ X | f (x) > 0} are disjoint and contain A and B respectively. Consequently, the space X is normal. Proposition 1.5.4 Every compact Hausdorff space is normal. Proof Let A and B be disjoint closed subsets of a compact Hausdorff space X . We want to show that there exist disjoint open subsets U and V of X such that A ⊂ U and B ⊂ V . Consider first the case where B is reduced to a single point y. As X is Hausdorff, we can find, for each x ∈ A, disjoint open subsets Ux and Vx such that x ∈ Ux and y ∈ Vx . By compactness of A, there exists a finite sequence x1 , . . . , xn of points in A such that A ⊂ Ux1 ∪ · · · ∪ Uxn . Then the open subsets U := Ux1 ∪ · · · ∪ Uxn and V := Vx1 ∩ · · · ∩ Vxn have the required properties. Let us now treat the general case. By the first part of the proof, we can find, for each y ∈ B, disjoint open subsets U y and Vy such that A ⊂ U y and {y} ⊂ Vy . By compactness of B, there exists a finite sequence y1 , . . . , yn of points in B such that B ⊂ Vy1 ∪ · · · ∪ Vyn . Then the open subsets U := U y1 ∩ · · · ∩ U yn and V := Vy1 ∪ · · · ∪ Vyn have the required properties. Remark 1.5.5 There exist compact accessible spaces that are not normal. Indeed, consider an infinite set X equipped with its cofinite topology. Then X is compact and accessible but not Hausdorff (cf. Example 1.3.2). Therefore, X is not normal since every normal accessible space is Hausdorff. Let us note that a subspace of a normal space may fail to be normal. In Sect. 5.4, we shall give an example of a compact Hausdorff (and hence normal) space containing an open subset that is not normal. However, we have the following result. Proposition 1.5.6 Every closed subset of a normal space is normal. Proof Let X be a normal space and F ⊂ X a closed subset. Let A and B be disjoint closed subsets of F. As F is closed in X , the sets A and B are closed in X . Since X is normal, we can find disjoint open subsets U and V of X such that A ⊂ U and B ⊂ V . Then the sets U ∩ F and V ∩ F are disjoint, open in F, and contain A and B respectively. Consequently, the space F is normal. Remark 1.5.7 In Sect. 5.5, we shall give an example of a normal Hausdorff space X such that X × X is not normal. Thus, the Cartesian product of two normal spaces is not necessarily normal.

14

1 Topological Dimension

1.6 Topological Dimension of Normal Spaces Two families of sets (E i )i∈I and (Fi )i∈I , with common indexed set I , are called combinatorially equivalent if one has

E i = ∅ ⇐⇒

i∈J

Fi = ∅,

i∈J

for every subset J ⊂ I . Remark 1.6.1 If α = (E i )i∈I and β = (Fi )i∈I are combinatorially equivalent families of sets and γ = (G i )i∈I is a family of sets such that E i ⊂ G i ⊂ Fi for all i ∈ I , then γ is combinatorially equivalent to α and β. Remark 1.6.2 If α = (Ai )i∈I and β = (Bi )i∈I are families of subsets of a set X that are combinatorially equivalent, then one has ord(α) = ord(β). Proposition 1.6.3 Let X be a normal space. Let (Fi )i∈I be a finite family of closed subsets of X and (Ui )i∈I a family of open subsets of X such that Fi ⊂ Ui for all i ∈ I . Then there exists a family (Vi )i∈I of open subsets of X satisfying the following conditions: (i) one has Fi ⊂ Vi ⊂ Vi ⊂ Ui for all i ∈ I ; (ii) the families(Fi )i∈I , (Vi )i∈I and (Vi )i∈I are combinatorially equivalent. Proof We can assume I = {1, . . . , n}. Let us set α = (F1 , . . . , Fn ). Let us show the existence, for every k ∈ {0, 1, . . . , n}, of a family (k)

α(k) = (A1 , . . . , A(k) n ) of subsets of X with the following properties: (k)

(C1) for every i ≤ k the set Ai

is open in X and one has (k)

Fi ⊂ Ai

(k)

⊂ Ai

⊂ Ui .

(C2) for every i ≥ k + 1, one has Ai(k) = Fi . (C3) the families of sets α and α(k) , where (k)

(k)

α(k) := (A1 , . . . , An ), are combinatorially equivalent.

1.6 Topological Dimension of Normal Spaces

15

Then the family α(n) will have the required properties, i.e., it will suffice to take (n) Vi := Ai for all i ∈ {1, . . . , n}. We prove the existence of α(k) by induction on k. For k = 0, we take α(0) := α, that is, Ai(0) := Fi for all i ∈ {1, . . . , n}. Then α(0) trivially satisfies (C1), (C2) and (C3). Suppose now that the family α(k−1) has already been constructed for some k ≤ n. We then define the family α(k) in the following way. For each i ∈ {1, . . . , n} such that i = k, we take Ai(k) := Ai(k−1) . (k)

It remains only to define Ak . Let us denote by E the set consisting of all subsets J ⊂ {1, . . . , n}\{k} such that Fk ∩

(k−1) Ai

= ∅.

i∈J

The set =

J ∈E

(k−1) Ai

i∈J

is closed in X since it is a finite union of closed subsets. We have Fk ⊂ X \ by definition of J . As X is normal, it follows from Proposition 1.5.2 that we can find an open subset W ⊂ X such that Fk ⊂ W ⊂ W ⊂ (X \) ∩ Uk . Let us take A(k) k := W . Then the family (k)

α(k) := (A1 , . . . , A(k) n ) clearly satisfies conditions (C1) and (C2). Let us show that the families α(k−1) (k−1)

and α(k) are combinatorially equivalent. As Ai Ak(k−1)

A(k) k

(k)

= Ai

for all i = k and

= Fk ⊂ by construction, it suffices to verify that, for every subset J ⊂ {1, . . . , n}\{k} such that (k−1) Ak

∩

(k−1) Ai

= ∅,

i∈J

one has (k) Ak

∩

i∈J

(k) Ai

= ∅.

16

1 Topological Dimension (k−1)

Since Ak

(k−1)

= Fk and Ai

(k)

= Ai

(k) Ak

∩

for all i = k, we have to check that

(k−1) Ai

=∅

i∈J (k)

for all J ∈ E. But this immediately follows from the fact that Ak = W ⊂ X \. We deduce that the families α(k−1) and α(k) are combinatorially equivalent. Conse quently, the family α(k) also satisfies (C3). Corollary 1.6.4 Let X be a normal space and let (Ui )i∈I be a finite open cover of X . Then there exists an open cover (Vi )i∈I of X such that Vi ⊂ Ui for all i ∈ I .

Proof The subsets Fi = X \Ui are closed in X and satisfy i∈I Fi = ∅. By Proposition 1.6.3, there exist open subsets Wi ⊂ X , i ∈ I , such that Fi ⊂ Wi and

i∈I Wi = ∅. The subsets Vi = X \Wi are open in X and cover X . On the other hand, we have Vi ⊂ X \Wi . As X \Wi is closed in X , it follows that Vi ⊂ X \Wi ⊂ X \Fi = Ui . Therefore, the subsets Vi have the required properties.

Proposition 1.6.5 Let X be a normal space and let α be a finite open cover of X . Then one has D(α) = min ord(γ), γ

where γ runs over all finite closed covers of X that are finer than α. Proof Suppose that α = (Ui )i∈I . Let γ = (F j ) j∈J be a finite closed cover of X that is finer than α. This means that there exists a map ϕ : J → I such that F j ⊂ Uϕ( j) for all j ∈ J . By Proposition 1.6.3, there exists a family β = (V j ) j∈J of open subsets of X that is combinatorially equivalent to γ and satisfies F j ⊂ V j ⊂ Uϕ( j)

(1.6.1)

for all j ∈ J . From (1.6.1), we deduce that β is a finite open cover of X that is finer than α. This implies D(α) ≤ ord(β) by definition of D(α). As ord(γ) = ord(β) since the covers γ and β are combinatorially equivalent, this shows that D(α) ≤ ord(γ). Conversely, suppose now that β = (V j ) j∈J is a finite open cover of X that is finer than α and satisfies D(α) = ord(β). By Corollary 1.6.4, there exists a closed cover γ = (F j ) j∈J of X such that F j ⊂ V j for all j ∈ J . Such a cover γ if finer than β and hence finer than α. Moreover, it satisfies ord x (γ) ≤ ord x (β) for all x ∈ X . It follows that ord(γ) ≤ ord(β) = D(α). Corollary 1.6.6 Let X be a normal space and let n ∈ N. Then the following conditions are equivalent:

1.6 Topological Dimension of Normal Spaces

17

(a) dim(X ) ≤ n; (b) for every finite open cover α of X , there exists a finite closed cover β α such that ord(β) ≤ n. Proof This is an immediate consequence of Proposition 1.6.5 since, by definition, dim(X ) = sup D(α), α

where α runs over all finite open covers of X .

Corollary 1.6.7 Let X be a compact metric space and let n ∈ N. Then the following conditions are equivalent: (a) dim(X ) ≤ n; (b) for every ε > 0, there exists a finite closed cover α of X such that mesh(α) ≤ ε and ord(α) ≤ n; (c) there exists a sequence (αk )k∈N of finite closed covers of X such that limk→∞ mesh(αk ) = 0 and ord(αk ) ≤ n for all k ∈ N. Proof Conditions (b) and (c) are clearly equivalent. Let ε > 0. If dim(X ) ≤ n, it follows from Proposition 1.4.4 that there exists a finite open cover β of X with ord(β) ≤ n and mesh(β) ≤ ε. We then have D(β) ≤ ord(β) ≤ n. Since X is normal, we deduce from Proposition 1.6.5 that there exists a closed cover α of X such that α β and ord(α) ≤ n. This shows that (a) implies (b). Conversely, suppose (b). Let α be a finite open cover of X . Let λ > 0 be a Lebesgue number of α. Since condition (b) is satisfied, there exists a finite closed cover β of X such that mesh(β) ≤ λ and ord(β) ≤ n. This implies β α and hence D(α) ≤ n by Proposition 1.6.5. Consequently, we have that dim(X ) ≤ n. This shows that (b) implies (a).

1.7 The Countable Union Theorem The following result is a generalization of Corollary 1.2.6. Theorem 1.7.1 Let X be a normal space and let (Fk )k∈N be a sequence of closed subsets of X such that X = k∈N Fk . Then one has dim(X ) = sup dim(Fk ). k∈N

(1.7.1)

Proof We have dim(X ) ≥ dim(Fk ) for all k ∈ N by Proposition 1.2.1. Therefore, it suffices to prove that (1.7.2) dim(X ) ≤ sup dim(Fk ). k∈N

18

1 Topological Dimension

Consider, for each k ∈ N, the closed subset Ak ⊂ X defined by Ak :=

Fs .

0≤s≤k

We have that dim(Ak ) = max dim(Fs ), 0≤s≤k

(1.7.3)

for all k ∈ N, by Corollary 1.2.6. Suppose now that α = (Ui )i∈I is a finite open cover of X . Let us inductively (k) construct a sequence β (k) = (Vi )i∈I , k ∈ N, of open covers of X satisfying the following conditions: (C1) Vi(0) = Ui for all i ∈ I , (k)

(k−1)

(C2) Vi ⊂ Vi for all i ∈ I and k ≥ 1, (C3) ord x (β (k) ) ≤ dim(Ak ) for all x ∈ Ak and k ≥ 1. We first take β (0) = α so that (C1) is satisfied. Suppose now that the covers β (0) , β (1) , . . . , β (k−1) have already been constructed for some integer k ≥ 1. As X (k−1) is a normal space, we can apply Corollary 1.6.4 to the cover β (k−1) = (Vi )i∈I . (k−1) for We deduce that there exists an open cover (Wi )i∈I of X such that Wi ⊂ Vi (k) (k) all i ∈ I . By Lemma 1.2.4, there exists an open cover β = (Vi )i∈I such that (k) Vi ⊂ Wi for all i ∈ I and ord x (β (k) ) ≤ dim(Ak ) for all x ∈ Ak . We then have (k)

Vi

(k−1)

⊂ Wi ⊂ Vi

for all i ∈ I . We deduce that the cover β (k) satisfies (C2) and (C3). Consider now the family γ = (L i )i∈I of closed subsets of X defined by L i :=

(k)

Vi

k∈N

for all i ∈ I . Let x be a point in X . As I is a finite set, there exists an index i 0 ∈ I such that (k) x ∈ Vi0 for infinitely many k ∈ N. By using (C2), we deduce that x ∈ L i0 . This (k)

shows that γ covers X . On the other hand, Condition (C2) implies that L i ⊂ Vi for all i ∈ I and k ≥ 0. Therefore, we have ord x (γ) ≤ ord x (β (k) ) for all k ≥ 0. As the sets Fk cover X , there exists an integer k0 = k0 (x) ≥ 1 such that x ∈ Ak0 . By Condition (C3), this implies ord x (γ) ≤ ord x (β (k0 ) ) ≤ dim(Ak0 ).

1.7 The Countable Union Theorem

19

By using (1.7.3), we deduce that ord(γ) ≤ sup dim(Ak ) = sup dim(Fk ). k∈N

k∈N

(0)

As L i ⊂ Vi = Ui for all i ∈ I , the finite closed cover γ is finer than α. By applying Proposition 1.6.5, we obtain D(α) ≤ ord(γ) ≤ sup dim(Fk ). k∈N

Since dim(X ) = supα D(α), this gives us (1.7.2).

Corollary 1.7.2 Every countable normal Hausdorff space X = ∅ has topological dimension dim(X ) = 0. Proof Let X ={x0 , x1 , x2 , . . . } be a non-empty countable normal Hausdorff space. We have X = k∈N Fk , where Fk = {xk } for all k ∈ N. It follows that dim(X ) = 0 since each Fk is closed in X with topological dimension dim(Fk ) = 0. As every metrizable space is normal by Proposition 1.5.3, we get the following result. Corollary 1.7.3 Every countable metrizable space X = ∅ has topological dimension dim(X ) = 0. Example 1.7.4 For every integer n ≥ 0, the space of rationals n-tuples Qn ⊂ Rn is metrizable and countable. Consequently, we have dim(Qn ) = 0. As every compact Hausdorff space is normal by Proposition 1.5.4, we also obtain the following. Corollary 1.7.5 Every countable compact Hausdorff space X = ∅ has topological dimension dim(X ) = 0. Remark 1.7.6 There exist countable compact accessible spaces with non-zero topological dimension. Indeed, if X is an infinite countable set equipped with its cofinite topology, then X is compact and accessible but dim(X ) = ∞ (see Examples 1.1.12 and 1.3.2). This also shows that Theorem 1.7.1 becomes false if we remove the normality hypothesis. An example of an infinite countable Hausdorff space with non-zero topological dimension will be given in Sect. 5.3. Corollary 1.7.7 The real line R has topological dimension dim(R) = 1. Proof It suffices to observe that R=

∞

Fk ,

k=1

where Fk = [−k, k], since dim(Fk ) = 1 for all k ≥ 1 by Proposition 1.4.5.

20

1 Topological Dimension

Remark 1.7.8 More generally, we shall prove in Corollary 3.5.7 that dim(Rn ) = n for every integer n ≥ 0.

1.8 Topological Dimension of Subsets of Metrizable Spaces In this section, we use the countable union theorem of the previous section (Theorem 1.7.1) to show that every subset Y of a metrizable space X satisfies dim(Y ) ≤ dim(X ). Lemma 1.8.1 Let X be a topological space. Suppose that every open subset ⊂ X satisfies dim() ≤ dim(X ). Then every subset Y ⊂ X satisfies dim(Y ) ≤ dim(X ). Proof Let Y ⊂ X . Let α = (Ui )i∈I be a finite open cover of Y . Then, for each i ∈ I , we can find an open subset Vi of X such that Ui = Y ∩ Vi . Consider the open subset of X defined by := i∈I Vi . As β := (Vi )i∈I is a finite open cover of , there exists a finite open cover γ = (W j ) j∈J of such that γ β and ord(γ) ≤ dim(). This implies ord(γ) ≤ dim(X ) since dim() ≤ dim(X ) by our hypothesis. Then the family δ := (Y ∩ W j ) j∈J is a finite open cover of Y that is finer than α and satisfies ord(δ) ≤ ord(γ) ≤ dim(X ). We deduce that D(α) ≤ dim(X ). Consequently, we have dim(Y ) = sup D(α) ≤ dim(X ). Let X be a topological space. A subset A ⊂ X is called an Fσ -set (resp. a G δ -set) if A is the union of some countable family of closed subsets of X (resp. the intersection of some countable family of open subsets of X ). Note that a subset A ⊂ X is an Fσ -set if and only if its complement X \ A is a G δ -set. Lemma 1.8.2 Every open subset of a metrizable space is an Fσ -set. Proof Let be an open subset of a metric space X . We may assume = X . Consider the continuous map f : X → R defined by f (x) := dist(x, X \). As X \ is closed in X , we have that f (x) = 0 if and only if x ∈ X \. It follows that = k∈N Fk , where Fk := f −1 ([

1 1 , ∞)) = {x ∈ X | dist(x, X \) ≥ }. k+1 k+1

As Fk is closed in X for all k ∈ N by continuity of f , this shows that is an Fσ -set. Theorem 1.8.3 Let X be a metrizable space and Y ⊂ X . Then one has dim(Y ) ≤ dim(X ).

1.8 Topological Dimension of Subsets of Metrizable Spaces

21

Proof By Lemma 1.8.1, it suffices to prove that dim() ≤ dim(X ) for every open subset of X . So let be an open subset of X . By Lemma 1.8.2, we can find a sequence (Fk )k∈N of closed subsets of X such that = k∈N Fk . The sets Fk are closed in . On the other hand, the space is normal since every subspace of a metrizable space is metrizable and hence normal. By applying Theorem 1.7.1, we obtain dim() = sup dim(Fk ). k∈N

As dim(Fk ) ≤ dim(X ) for all k ∈ N by Proposition 1.2.1, we conclude that dim() ≤ dim(X ). Remark 1.8.4 It may happen that dim(Y ) > dim(X ) when Y is a subset of a normal Hausdorff space X . Indeed, in Sect. 5.4, we will give an example of a compact Hausdorff (and hence normal) space X with dim(X ) = 0 containing an open subset of positive dimension.

Notes Covering dimension is one among many other invariants that were introduced by mathematicians all along the twentieth century in order to give a precise definition for the intuitive notion of dimension in the category of topological spaces. The branch of general topology that studies these invariants is known as “dimension theory”. This is also the title of the most famous monograph devoted to the subject, namely the book by Hurewicz and Wallman [50], which was first published in 1941. There are several other excellent books entirely devoted to dimension theory, e.g., [9, 33, 79, 80, 86]. The reader interested in the history of the developments of dimension theory is referred to [7, 8, 33, 56, 57, 92]. ˇ The covering dimension dim(X ) was introduced by Cech [111]. Its definition was directly inspired by a topological characterization of the dimension of the ncube [0, 1]n formulated by Lebesgue [66, 67] (see Lemma 3.5.2). The idea of using induction for defining the dimension of a topological space was popularized by Poincaré (see for example [89, p. 73]). This approach led in particular to the definition of the small inductive dimension ind(X ) and of the large inductive dimension Ind(X ). The small inductive dimension ind(X ) ∈ {−1} ∪ N ∪ {∞} of a topological space X , also called the Menger-Urysohn dimension, is inductively defined by the following conditions: (1) ind(X ) = −1 if and only if X = ∅, (2) ind(X ) ≤ n if and only if X admits a base of open subsets B such that ind(U \U ) ≤ n − 1 for all U ∈ B. The large inductive dimension Ind(X ) ∈ {−1} ∪ N ∪ {∞}, ˇ also called the Brouwer-Cech dimension, is defined by: (1) Ind(X ) = −1 if and only if X = ∅, (2) Ind(X ) ≤ n if and only if, for every pair of disjoint closed subsets F and G of X , there exist disjoint open subsets U and V of X such that F ⊂ U , G ⊂ V and Ind(X \(U ∪ V )) ≤ n − 1. An easy induction shows that every accessible space X satisfies ind(X ) ≤ Ind(X ). Urysohn’s theorem asserts

22

1 Topological Dimension

that dim(X ) = ind(X ) = Ind(X ) for every separable metrizable space X (see for example [50]). Katˇetov [54, 55] and independently Morita [78] proved that one has ind(X ) ≤ dim(X ) = Ind(X ) for every metrizable space X . The question whether every metrizable space X satisfies dim(X ) = ind(X ) remained open for a long time (cf. [7, p. 3]). Finally, it was answered in the negative in [96, 98] by Roy who provided an example of a metrizable space X with ind(X ) = 0 and dim(X ) = 1. It was shown by Pasynkov [84] that the equalities dim(X ) = ind(X ) = Ind(X ) remain true when X is the underlying space of a locally compact Hausdorff topological group. A theorem of Alexandroff [6] asserts that every compact Hausdorff space X satisfies dim(X ) ≤ ind(X ) ≤ Ind(X ). Filippov [36] gave an example of a compact Hausdorff space X with dim(X ) = ind(X ) = 2 and Ind(X ) = 3. When X is a normal space, one always has dim(X ) ≤ Ind(X ) (see for example [33, 79, 86]) but the inequality may be strict. In [79, p. 114], Nagami gives an example of a normal Hausdorff space X such that ind(X ) = 0, dim(X ) = 1, and Ind(X ) = 2. The notion of a normal space goes back to the work of Vietoris [112] (see [94, p. 1233]) and Tietze [105]. However, the main results about general properties of normal spaces are due to Urysohn [109]. ˇ Theorems 1.7.1 and 1.8.3 were obtained by Cech [111]. ˇ In [110], Cech introduced the following definition. A topological space X is called perfectly normal if X is normal and every open subset of X is an Fσ -set. For example, every metrizable space is perfectly normal by Proposition 1.5.3 and Lemma 1.8.2. It turns out that every perfectly normal space X is completely normal, that is, every subset Y ⊂ X is normal (see [18 exerc. 7, 9 and 11 p. IX. 102–103], [64, 111]). Thus, the proof of Theorem 1.8.3 can be extended to perfectly normal spaces. Consequently, every subset Y of a perfectly normal space X satisfies dim(Y ) ≤ dim(X ) [111, par. 28]. Alexandroff (see [8, p. 28]) conjectured that dim(Y ) ≤ dim(X ) whenever Y is a normal subspace of a normal space X . This conjecture was disproved by Dowker [30] who gave an example of a normal Hausdorff space X with dim(X ) = 0 containing a normal open subset Y such that dim(Y ) = 1. The idea of finding a homological interpretation of the dimension of a topological space was developed in the work of Alexandroff [4, 5] in the late 1920s. It subsequently led to the investigation of various notions of homological and cohomological dimension (see the books [9, 50, 79, Appendix by Kodama] and the survey papers [31, 32, 65]). Given a topological space X and an abelian group G, the cohomological dimension cdim G (X ) is defined as being the smallest integer n ≥ −1 such that ˇ n+1 (X, A; G) = 0 for all closed subsets A ⊂ X , or ∞ if there is no such integers. H ˇ ˇ ∗ denotes relative Cech cohomology. It was shown by Alexandroff that every Here H compact metrizable space X with dim(X ) < ∞ satisfies dim(X ) = cdimZ (X ). In the first International Topological Conference held in Moscow in September 1935, Alexandroff asked if this equality remains true in the case when dim(X ) = ∞. This question was answered in the negative in the late 1980s by Dranishnikov [31] who proved, by using methods from K-theory, the existence of a compact metrizable space X with topological dimension dim(X ) = ∞ and integral cohomological dimension cdimZ (X ) = 3.

Exercises

23

Exercises 1.1 Show that every finite topological space X satisfies dim(X ) < ∞. 1.2 Show that the topological space X described in Example 1.1.11 is connected but not accessible. 1.3 Let X be an infinite set and x0 ∈ X . The set X is equipped with the topology for which the open subsets are ∅ and all the subsets of X containing x0 . Show that dim(X ) = ∞. 1.4 Let X be the topological space whose underlying set is R and whose open subsets are ∅, R, and all the intervals of the form (a, +∞), where a ∈ R. Show that X is connected and that dim(X ) = 0. 1.5 Let X be a non-empty set and π a partition of X . The set X is equipped with the topology for which the open subsets are ∅ and all the subsets of X that can be written as a union of elements of π. Show that dim(X ) = 0. 1.6 Show that every finite accessible topological space is discrete. 1.7 Let X and Y be topological spaces with Y accessible and non-empty. Show that dim(X × Y ) ≥ dim(X ). 1.8 Let α = {U, V } be the open cover of R defined by U :=

(n, n + 1) and V :=

n∈Z

1.9 1.10

1.11

1.12 1.13 1.14 1.15

(n −

n∈Z

1 1 ,n + ). |n| + 1 |n| + 1

Show that α admits no Lebesgue numbers, i.e., for every λ > 0, there exists a subset Y ⊂ R such that diam(Y ) ≤ λ that is contained in no element of α. Let S1 := {(x, y) ∈ R2 | x 2 + y 2 = 1} denote the unit circle in R2 . Show that dim(S1 ) = 1. Construct, for every ε > 0, an open cover α of R2 with order ord(α) = 2 and Euclidean mesh mesh(α) ≤ ε. Deduce that every compact subset X ⊂ R2 satisfies dim(X ) ≤ 2. Deduce from the previous exercise and the countable union theorem (Theorem 1.7.1) that dim(R2 ) ≤ 2. More generally, show that dim(Rn ) ≤ n for every n ∈ N. (The fact that dim(Rn ) = n will be proved in Corollary 3.5.7 below). Deduce from the previous exercise that one has dim(Y ) ≤ n for every subset Y ⊂ Rn . Show that the topological space X of Example 1.1.11 is normal if and only if n = 0. Let X be a normal Hausdorff space. Show that any two distinct points of X admit disjoint closed neighborhoods. Let T denote the set consisting of all subsets of R of the form U \C, where U is an open subset of R for the usual topology and C ⊂ U is a countable subset.

24

1 Topological Dimension

(a) Show that T is the set of open sets of a topology on R. Let X denote the topological space whose underlying set is R and whose set of open subsets is T . (b) Show that any two distinct points of X admit disjoint closed neighborhoods. (c) Show that X is not normal. Hint: consider the sets A := {0} and B := {1/n | n ≥ 1}. 1.16 Let X ⊂ R. Show that the following conditions are equivalent: (1) dim(X ) = 1, (2) X contains a subset homeomorphic to the unit segment [0, 1], (3) the interior of X in R is not empty. 1.17 Let Y := [0, 1] denote the unit segment in R and let X := Y ∪ {x0 } be the set / Y . Equip X with the topology obtained from Y by adjoining an element x0 ∈ for which the open subsets are X and all the subsets ⊂ Y such that is an open subset for the usual topology on Y . (a) (b) (c) (d)

Show that X is not accessible. Show that X is compact and connected. Show that every subspace of X is normal. Show that dim(X ) = 0 but dim(Y ) = 1.

1.18 Let X be a compact metric space. Let (αk )k∈N be a sequence of finite open covers of X such that limk→∞ mesh(αk ) = 0. Show that dim(X ) = limk→∞ D(αk ). 1.19 Let T be a triangle in the Euclidean plane (i.e., the convex hull of three noncollinear points in R2 ). The middle-triangle of T is the interior in R2 of the

g2

f2

f3

d2 Fig. 1.3 Construction of a Sierpinski triangle

Exercises

25

triangle whose vertices are the midpoints of the sides of T . We inductively construct a decreasing sequence (K n )n∈N of subsets of R2 in the following way. We start by setting K 0 := T . Then we define K 1 as being the set obtained from K 0 = T by removing its middle-triangle. Thus, K 1 is the union of three triangles that are the images of T by the homotheties of ratio 1/2 centered at each of the vertices of T . More generally, assuming that K n has already been constructed and is the union of 3n triangles that are all pairwise disjoint except at some of their vertices, we define K n+1 as being the set obtained from K n by removing all the middle-triangles

of these 3n triangles. The Sierpinski triangle associated with T is the set S := n∈N K n (see Fig. 1.3). (a) Show that the homeomorphism type of S does not depend on the initial choice of T . (b) Show that S is a connected compact subset of R2 . (c) Show that dim(S) = 1. Hint: observe that the construction yields a sequence (βn )n∈N of finite closed covers of S with ord(βn ) = 1 and Euclidean mesh mesh(βn ) =

diam(T ) 2n

and then apply Corollary 1.6.7 to get dim(S) ≤ 1.

Chapter 2

Zero-Dimensional Spaces

This chapter is devoted to 0-dimensional topological spaces. It follows from the definition of topological dimension given in Chap. 1 that a zero-dimensional topological space admits arbitrarily fine open partitions. As every element of an open partition is a clopen subset, i.e., a subset that is both closed and open, this suggests that any zero-dimensional space must contain many clopen subsets and hence be very disconnected since the abundance of clopen subsets reflects the discontinuous nature of a topological space. We shall study the relationship between the class of zerodimensional topological spaces and other classes of highly-disconnected topological spaces such as the class of scattered spaces, the class of totally disconnected spaces, and the class of totally separated spaces.

2.1 The Cantor Set In this section, we first describe the construction of the Cantor set, which is a fundamental example of a compact metrizable space with zero topological dimension. Let a and b be real numbers such that a < b. The open interval b−a 2a + b a + 2b b−a a+ ,b − , = 3 3 3 3 is called the middle third of the segment [a, b]. We denote by T ([a, b]) the set obtained by deleting from the segment [a, b] its middle third. Thus, we have a + 2b 2a + b b−a b−a , b = a, ,b . ∪ b− ∪ T ([a, b]) := a, a + 3 3 3 3

© Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_2

27

28

2 Zero-Dimensional Spaces

More generally, for every subset A ⊂ R which is the union of a finite family ([ai , bi ])1≤i≤k of pairwise disjoint segments, we set T (A) :=

k

T ([ai , bi ]).

i=1

Let us inductively define a decreasing sequence (K n )n∈N of closed subsets of [0, 1] by setting K 0 := [0, 1], K n+1 := T (K n ) for all n ∈ N. We therefore have 2 1 ,1 , ∪ K 1 = 0, 3 3 2 1 2 7 8 1 , , ,1 , ∪ ∪ ∪ K 2 = 0, 9 9 3 3 9 9 2 1 2 7 8 1 1 , , , ∪ ∪ ∪ K 3 = 0, 27 27 9 9 27 27 3 20 7 8 25 26 2 19 , , , , 1 , etc. ∪ ∪ ∪ ∪ 3 27 27 9 9 27 27 Observe that the set K n is the union of 2n pairwise disjoint segments of length 1/3n . These segments are the connected components of K n (see Fig. 2.1). The set Kn K := n∈N

is called the Cantor ternary set or simply the Cantor set. A topological space that is homeomorphic to the Cantor ternary set K is called a Cantor space. Proposition 2.1.1 The Cantor set K is a compact subset of R with empty interior. Proof As the sets K n are closed in [0, 1], the Cantor set is closed in [0, 1] and hence compact.

Fig. 2.1 Construction of the Cantor set

2.1 The Cantor Set

29

Let I be an interval of R such that I ⊂ K . The fact that I is connected implies that, for each n ∈ N, the set I is contained in one of the 2n connected components of K n . We deduce that the length of I is smaller than or equal to 1/3n for all n ∈ N. As 1/3n tends to 0 as n goes to infinity, it follows that I has zero length, i.e., is either empty or reduced to a single point. This shows that K has empty interior. Proposition 2.1.2 The Cantor set K has topological dimension dim(K ) = 0. Proof The set K n is the disjoint union of 2n segments n (i), 1 ≤i ≤ 2n , which are clopen in K n . Let us set Un (i) := K ∩ n (i). The family αn := Un (i) 1≤i≤2n is a finite open partition of K . Therefore, we have ord(αn ) = 0. As mesh(αn ) = 1/3n tends to 0 as n goes to infinity, we deduce that dim(K ) = 0 by applying Proposition 1.4.4 (observe that the set K is not empty since we clearly have 0 ∈ K ). Recall that every real number x ∈ [0, 1] admits a ternary expansion, that is, a sequence (u k )k∈N ∈ {0, 1, 2}N such that x=

∞

uk . 3k+1 k=0

We will also write this equality under the form x = 0, u 0 u 1 u 2 · · · u k · · ·. When x is not a triadic rational number of the form n/3m with n and m integers satisfying 1 ≤ n ≤ 3m − 1, such an expansion is unique. In the case when x = n/3m with n and m integers such that 1 ≤ n ≤ 3m − 1, the number x admits two ternary expansions: a first one, called the proper ternary expansion of x, whose terms are eventually equal to 0 and another one, called the unproper ternary expansion of x, whose terms are eventually equal to 2. For example, we have 1 = 0, 02020202 . . . 4 and

7 = 0, 210000 . . . = 0, 202222 . . .. 9

The set K n consists of all numbers x ∈ [0, 1] that admit a ternary expansion (u k )k∈N such that u k ∈ {0, 2} for all k ≤ n − 1. We deduce that the Cantor set K is the set consisting of the numbers x ∈ [0, 1] that admit a ternary expansion whose terms all belong to the set {0, 2}. Thus, the ternary expansions given above show that both 1/4 and 7/9 belong to K .

30

2 Zero-Dimensional Spaces

Proposition 2.1.3 The map ϕ : {0, 1}N → K defined by ϕ(u) :=

∞

2u k 3k+1 k=0

for all u = (u k ) ∈ {0, 1}N is a homeomorphism from the product space {0, 1}N onto the Cantor set K . Proof The fact that the map ϕ is well defined and bijective follows from the previous observations. Let us fix a sequence u ∈ {0, 1}N . For each integer n ≥ 0, the set Vn (u) ⊂ {0, 1}N consisting of all sequences v such that vk = u k for all k ≤ n is an open neighborhood of u. For all v ∈ Vn (u), we have that |ϕ(u) − ϕ(v)| ≤

∞

k=n+1

2 1 = n+1 . 3k+1 3

Since 1/3n+1 tends to 0 as n goes to infinity, we deduce that ϕ is continuous. The space {0, 1}N is compact as it is a product of compact spaces. Consequently, ϕ is a homeomorphism. Corollary 2.1.4 The Cantor set is uncountable.

Let X be a topological space. A point x ∈ X is called isolated if the singleton set {x} is open in X . A topological space is called perfect if it contains no isolated points. Corollary 2.1.5 The Cantor set is perfect. Proof Let u ∈ {0, 1}N . Consider the open subsets Vn (u) := {v ∈ {0, 1}N | vk = u k for all k ≤ n} ⊂ {0, 1}N . By definition of the product topology, every neighborhood of u in {0, 1}N contains the sets Vn (u) for n large enough. As the set Vn (u) is infinite for every n, we deduce that u is not isolated. This shows that the space {0, 1}N is perfect. As K is homeomorphic to {0, 1}N , it is also perfect.

2.2 Scattered Spaces In this section, we introduce the class of scattered spaces. We prove that an accessible topological space X is scattered if and only if there exists a set E such that X is homeomorphic to a subspace of the product space {0, 1} E .

2.2 Scattered Spaces

31

Let X be a topological space. A base of the topological space X is a set B of open subsets of X such that every open subset of X can be written as a union of elements of B. A set N of neighborhoods of a point x ∈ X is called a neighborhood base of x if, for every neighborhood V of x, there exists N ∈ N such that N ⊂ V . Observe that a set B of open subsets of X is a base of X if and only if, for every x ∈ X , the set Bx := {B ∈ B | x ∈ B} is a neighborhood base of the point x. If B is a base of a topological space X , then B satisfies the following two conditions: (B1) the elements of B cover X ; (B2) if B1 , B2 ∈ B and x ∈ B1 ∩ B2 , then there exists B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2 . Conversely, if X is a set and B is a set of subsets of X satisfying conditions (B1) and (B2) above, then there exists a unique topology on X admitting B as a base. Example 2.2.1 Let X be a metric space. Then the set consisting of all open balls B(x, 1/n), where x ∈ X and n ≥ 1 is an integer, is a base of X . Recall that a subset of a topological space X is said to be clopen if it is both open and closed in X . Note that the clopen subsets of a topological space are precisely the subsets with empty boundary. Definition 2.2.2 We say that a topological space X is scattered if it admits a base consisting of clopen subsets of X . A topological space X is scattered if and only if every point of X admits a neighborhood base consisting of clopen subsets. Example 2.2.3 Every set endowed with the discrete topology is scattered. Remark 2.2.4 A connected space X is scattered if and only if the topology on X is the trivial one. Note that a scattered space may fail to be accessible. For example, every set X equipped with the trivial topology is scattered. However, such a space X is not accessible as soon as X contains more than one point. Proposition 2.2.5 Every scattered accessible space is Hausdorff. Proof Let X be a scattered accessible space. Let x and y be distinct points in X . Since X is accessible, the set X \{y} is an open neighborhood of x. As X is scattered, there exists a clopen neighborhood V of x that is contained in X \{y}. The sets V and X \V are disjoint open subsets of X containing x and y respectively. This shows that X is Hausdorff.

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Proposition 2.2.6 Every subspace of a scattered space is itself scattered. Proof Let X be a scattered space and Y ⊂ X . If B is a base of X consisting of clopen subsets, then the sets Y ∩ B, where B ∈ B, are clopen in Y and form a base of Y . Consequently, Y is scattered. Proposition 2.2.7 Every product of scattered spaces is itself scattered. Proof Let (X i )i∈I be a family of scattered spaces and consider their direct product X := i∈I X i . Let Bi be a base of X i consisting of clopen subsets. We can assume X i ∈ Bi . Then the set i∈I Ui , where Ui ∈ Bi for all i ∈ I and Ui = X i for all but finitely many i ∈ I , are clopen in X and form a base for the product topology. Therefore X is scattered. Every open ball of the Euclidean space Rn is connected. Consequently, every scattered subset of Rn (n ≥ 1) has empty interior. For the subsets of R, the converse is also true: Proposition 2.2.8 Let X be a subset of the real line R. Then X is scattered if and only if it has empty interior. Proof We already observed that the condition is necessary. Let us show that it is also sufficient. Suppose that X has empty interior. Let x ∈ X and ε > 0. As X has empty interior, we can find real numbers a and b not in X such that x − ε < a < x < b < x + ε. Then the set V := (a, b) ∩ X = [a, b] ∩ X is a clopen neighborhood of x in X satisfying V ⊂ (x − ε, x + ε). This shows that X is scattered. By applying the preceding proposition, we see that the set of rational numbers Q, the set of irrational numbers R\Q, and the Cantor set K are all scattered. Proposition 2.2.9 Let X be an accessible space. Then the following conditions are equivalent: (a) the space X is scattered; (b) there exists a set E such that X is homeomorphic to a subset of the product space {0, 1} E . Proof Given a set E, the space {0, 1} E is a product of discrete spaces and hence scattered by Proposition 2.2.7. As every subset of a scattered space is itself scattered by Proposition 2.2.6, this shows that (b) implies (a). Conversely, suppose that X is a scattered space. Let E be a base of X consisting of clopen subsets. Consider the map ϕ : X → {0, 1} E defined by ϕ(x) = (χ B (x)) B∈E , where χ B : X → {0, 1} is the characteristic map of B. As B is clopen in X , the map χ B is continuous for each B ∈ E. It follows that ϕ is continuous. On the other hand, if x and y are distinct points in X , then X \{x} is an open neighborhood of y since X is accessible. Therefore, there exists a neighborhood B0 ∈ E of y such that B0 ⊂ X \{x}. This implies χ B0 (x) = χ B0 (y) and hence ϕ(x) = ϕ(y). We deduce that ϕ is injective. We have that ϕ(B) = ϕ(X )∩π B−1 (1), where π B : {0, 1} E → {0, 1}

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is the projection map onto the B-factor of {0, 1} E . This shows that ϕ(B) is open in ϕ(X ) for all B ∈ E. As E is a base of X , we deduce that the image by ϕ of every open subset of X is open in ϕ(X ). Consequently, ϕ induces a homeomorphism from X onto ϕ(X ). Therefore, the space X satisfies (b).

2.3 Scatteredness of Zero-Dimensional Spaces In this section, we give a characterization of 0-dimensional topological spaces. This characterization shows that every 0-dimensional accessible space is scattered. Theorem 2.3.1 Let X be a non-empty topological space. Then the following conditions are equivalent: (a) dim(X ) = 0; (b) for every pair of disjoint closed subsets A and B of X , there exist disjoint open subsets U and V of X such that X = U ∪ V , A ⊂ U and B ⊂ V ; (c) for every closed subset A of X and every open subset U of X such that A ⊂ U , there exists a clopen subset V of X such that A ⊂ V ⊂ U . Proof Suppose first that dim(X ) = 0. Let A and B be disjoint closed subsets of X . Consider the open cover α = {X \A, X \B}. As dim(X ) = 0, there exists a finite open partition β of X such that β α. Note that no element of β can meet both A and B. Denote by U the union of all the elements of β that meet A and let V := X \U . The sets U and V form an open partition of X . Moreover, we have that A ⊂ U and B ⊂ V . This shows that (a) implies (b). Let us show now that (b) implies (c). Suppose that X satisfies (b). Let A be a closed subset of X and U an open subset of X such that A ⊂ U . Then B := X \U is a closed subset that does not meet A. By (b), it follows that there exists a partition of X into two open subsets V and W such that A ⊂ V and B ⊂ W . Then the set V is a clopen subset of X and we have A ⊂ V ⊂ U . This shows that X satisfies (c). Finally, let us prove that (c) implies (a). Suppose that X satisfies (c). Let α = (Ui )i∈I be a finite open cover of X . As X satisfies (c), it follows from Proposition 1.5.2 that X is normal. By applying Corollary 1.6.4, we deduce that there exists a closed cover (Fi )i∈I of X such that Fi ⊂ Ui for all i ∈ I . Since X satisfies (c), we can find, for each i ∈ I , a clopen subset Vi of X such that Fi ⊂ Vi ⊂ Ui . Without loss of generality, we may assume that I = {1, . . . , n}. Consider the family β = (Wi )i∈I of subsets of X defined by W1 := V1 and Wi := Vi \(V1 ∪ · · · ∪ Vi−1 ) for all i ∈ {2, . . . , n}. Clearly β := (Wi )i∈I is an open partition of X . Moreover, we have that β α since Wi ⊂ Vi ⊂ Ui for all i ∈ I . This shows that dim(X ) = 0. Corollary 2.3.2 Every topological space X satisfying dim(X ) = 0 is normal.

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Proof A topological space X such that dim(X ) = 0 satisfies condition (b) in the preceding theorem and is therefore normal. Corollary 2.3.3 Every accessible topological space X satisfying dim(X ) = 0 is scattered. Proof Let X be an accessible space such that dim(X ) = 0. Let V be a neighborhood of a point x ∈ X . The singleton {x} is closed in X since X is accessible. As dim(X ) = 0, the space X satisfies condition (c) of the preceding theorem. Therefore, there exists a clopen subset U of X such that x ∈ U ⊂ V . Consequently, every point of X admits a neighborhood base consisting of clopen subsets of X . This shows that X is scattered. Corollary 2.3.4 If X is an accessible topological space such that dim(X ) = 0, then X is Hausdorff. Proof Every scattered accessible space is Hausdorff by Proposition 2.2.5.

Remark 2.3.5 Corollary 2.3.4 can also be deduced from Corollary 2.3.2 since, as already observed in Sect. 1.5, every normal accessible space is clearly Hausdorff. In Sect. 5.4, we shall give an example of a locally compact Hausdorff space that is scattered but not normal. Such a space has positive topological dimension by Corollary 2.3.2.

2.4 Lindelöf Spaces In this section, we introduce the class of Lindelöf spaces and we prove that every non-empty scattered Lindelöf space X has topological dimension dim(X ) = 0. Definition 2.4.1 A topological space X is called a Lindelöf space if every open cover of X admits a countable subcover. Example 2.4.2 Every countable topological space is Lindelöf. Indeed, suppose that X is a countable topological space. Let α = (Ui )i∈I be an open cover of X . Choose, for each x ∈ X , an index i(x) ∈ I such that x ∈ Ui(x) . Let J := {i(x) | x ∈ X }. Then β := (Ui )i∈J is a countable subcover of α. Example 2.4.3 Every compact space is Lindelöf. Indeed, by definition, a topological space X is compact if and only if every open cover of X admits a finite subcover. Example 2.4.4 Every topological space that is a union of a countable family of subsets that are Lindelöf (for the induced topology) is Lindelöf. In particular, every σ -compact space is Lindelöf (recall that a topological space is called σ -compact if it is the union of a countable family of compact subsets). Thus, the Euclidean space Rn is Lindelöf for any integer n ≥ 1 since it is σ -compact.

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Example 2.4.5 If an uncountable set X is endowed with its discrete topology, then X is not Lindelöf. Indeed, the open cover α := ({x})x∈X admits no countable subcovers. Note that X is metrizable (a metric inducing the topology on X is given by d(x, y) = 0 if x = y and d(x, y) = 1 otherwise) and locally compact. A subset of a Lindelöf space is not necessarily Lindelöf (see the example in Sect. 5.4). However, we have the following result. Proposition 2.4.6 Every closed subset of a Lindelöf space is itself Lindelöf. Proof Let X be a Lindelöf space and F a closed subset of X . Let α = (Ui )i∈I be an open cover of F. Then we can find, for each i ∈ I , an open subset Vi of X such that Ui = Vi ∩ F. As the family (Vi )i∈I ∪ {X \F} is an open cover of X and X is Lindelöf, there exists a countable subset J ⊂ I such that the family (V j ) j∈J ∪{X \F} covers X . Then the family (U j ) j∈J is a countable subcover of α. This shows that F is Lindelöf. Remark 2.4.7 The product of two Lindelöf spaces may fail to be Lindelöf (see Sect. 5.5). Definition 2.4.8 A topological space is said to be second-countable if it admits a countable base. For example, the Euclidean space Rn is second-countable since the open balls B(x, 1/m), where x ∈ Qn and m ≥ 1 is an integer, form a countable base of Rn . A topological space X is called first-countable if every point of X admits a countable neighborhood base. Clearly every second-countable topological space is also first-countable. On the other hand, a first-countable space is not necessarily secondcountable. For example, an uncountable set equipped with its discrete topology is first-countable but not second-countable. Proposition 2.4.9 Every subset of a second-countable topological space is itself second-countable. Proof If X is a topological space admitting a countable base B and Y ⊂ X , then the set consisting of all the subsets of the form Y ∩ B, where B runs over B, is clearly a countable base for Y . Proposition 2.4.10 Every countable product of second-countable spaces is itself second-countable. Proof Let (X i )i∈I be a countable family of second-countable spaces and consider their direct product X := X . Let Bi be a countable base of X i . We can assume i i∈I X i ∈ Bi . Then the sets i∈I Ui , where Ui ∈ Bi for all i ∈ I and Ui = X i for all but finitely many i ∈ I , form a countable base for the product topology. Therefore X is second-countable. Proposition 2.4.11 (Lindelöf) Every second-countable topological space is Lindelöf.

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Proof Let X be a topological space admitting a countable base B. Let α = (Ui )i∈I be an open cover of X . Denote by B the set consisting of all B ∈ B such that there exists i ∈ I satisfying B ⊂ Ui . Define a map ϕ : B → I by choosing, for each B ∈ B , an index ϕ(B) ∈ I such that B ⊂ Uϕ(B) . Then the image set J = ϕ(B ) ⊂ I is countable. Let x ∈ X . As α covers X , we can find an index i(x) ∈ I such that x ∈ Ui(x) . Since B is a base of X , there exists an open subset B(x) ∈ B such that x ∈ B(x) ⊂ Ui(x) . We have that B(x) ∈ B , by definition of B , and x ∈ B(x) ⊂ Uϕ(B(x)) . It follows that (Ui )i∈J is a countable cover of X . This shows that X is Lindelöf. Definition 2.4.12 A topological space is said to be separable if it admits a countable dense subset. Proposition 2.4.13 Every second-countable topological space is separable. Proof Let X be a topological space and B a base of X . Let us choose, for each B ∈ B with B = ∅, a point x B ∈ B and denote by Y the set consisting of all such points x B . Since B is a base for X , every non-empty open subset of X contains a point of Y . Consequently, Y is dense in X . If B is countable, then Y is also countable and hence X is separable. From Propositions 2.4.9, 2.4.11 and 2.4.13, we immediately deduce the following result. Corollary 2.4.14 Every subset of a second-countable space is separable and Lindelöf. In particular, every subset of the Euclidean space Rn is separable and Lindelöf. The following example shows that a separable compact Hausdorff space may fail to be first-countable. Example 2.4.15 Let X denote the set consisting of all maps from R into the unit segment [0, 1]. We equip X with the topology of pointwise convergence. Thus, the space X may be identified with the product space [0, 1]R and is a compact Hausdorff space by Tychonoff’s theorem. Let f ∈ X . By definition of the topology of pointwise convergence, for every ε > 0 and every finite subset A ⊂ R, the set V ( f, ε, A) := {g ∈ X | | f (x) − g(x)| < ε for all x ∈ A} is an open neighborhood of f . Moreover, the sets V ( f, ε, A), where ε > 0 and A ⊂ R is a finite subset, form a neighborhood base of f . Let D denote the subset of X consisting of all finite linear combinations with rational coefficients of characteristic maps of segments of R with rational endpoints. Clearly D is dense in X . As D is countable, this shows that X is separable. However, X is not first-countable. Otherwise, every f ∈ X would admit a countable neighborhood base Wn , n ∈ N. Then, for every n ∈ N, there would exist εn > 0 and a finite subset An ⊂ R such that V ( f, εn , An ) ⊂ Wn . The set E := n∈N An would be countable and hence we

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would have R\E = ∅. Taking a point x0 ∈ R\E, any map g : R → [0, 1] such that g(x0 ) = f (x0 ) and g(x) = f (x) for all x ∈ E would satisfy g ∈ Wn for all n ∈ N. As X is Hausdorff, this would imply g = f , which contradicts g(x0 ) = f (x0 ). Consequently, X is not first-countable and hence not second-countable either. Remark 2.4.16 The topological space in the preceding example is not metrizable. Indeed, every metrizable space is first-countable since, in a metric space X , every point x ∈ X admits a countable neighborhood base, e.g., the one formed by the open balls B(x, 1/n), n ≥ 1. Remark 2.4.17 The space X in Example 2.4.15 is Lindelöf since it is compact. In Sect. 5.5, we will describe a first-countable separable Lindelöf Hausdorff space S which is not second-countable (see Proposition 5.5.1 and Corollary 5.5.7). For metrizable spaces, we have the following equivalent conditions. Proposition 2.4.18 Let X be a metrizable space. Then the following conditions are equivalent: (a) (b) (c) (d)

X X X X

is second-countable; is Lindelöf; is separable; is homeomorphic to a subset of the Hilbert cube [0, 1]N .

Proof The fact that (a) implies (b) follows from Proposition 2.4.11. Let us fix a metric d on X compatible with its topology. Suppose (b). Given an integer n ≥ 1, consider the cover of X formed by the open balls B(x, 1/n), x ∈ X . As X is Lindelöf, there exists a countable subset Yn ⊂ X such that the balls B(y, 1/n), y ∈ Yn , cover X . The set Y := n≥1 Yn is countable and dense in X . Consequently, X is separable. This shows that (b) implies (c). The unit segment [0, 1] ⊂ R is second-countable. Thus, condition (d) implies (a) since any countable product of second-countable topological spaces is secondcountable by Proposition 2.4.10 and any subset of a second-countable space is second-countable by Proposition 2.4.9. To complete the proof, it suffices to show that (c) implies (d). Suppose (c). Let A = {an | n ∈ N} be a countable dense subset of X . After possibly replacing d(x, y) by the metric min(d(x, y), 1), which is also compatible with the topology on X , we can assume that diam(X ) ≤ 1. Consider the map F : X → [0, 1]N defined by F(x) = (d(x, an ))n∈N . The map F is continuous since all maps x → d(x, an ) are continuous. As every point of X is the limit of some sequence of points in A, it follows that F is injective (uniqueness of the limit in Hausdorff spaces). Let now x0 ∈ X and ε > 0. As A is dense in X , there exists an integer n 0 ≥ 0 such that d(x0 , an 0 ) < ε/2. Then the subset U ⊂ [0, 1]N consisting of all sequences (u n )n∈N ∈ [0, 1]N such that u n 0 < ε/2 is

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an open neighborhood of F(x0 ). If x ∈ X is such that F(x) ∈ U , then x satisfies d(x, an 0 ) < ε/2 and hence d(x, x0 ) ≤ d(x, an 0 ) + d(x0 , an 0 ) < ε, by applying the triangle inequality. Consequently, we have that F −1 (U ) ⊂ B(x0 , ε). We deduce that F induces a homeomorphism from X onto F(X ). This shows that X satisfies (d). As every compact space is Lindelöf, we immediately get the following: Corollary 2.4.19 Every compact metrizable space is second-countable and hence separable. It follows from Corollary 2.3.3 that every accessible topological space X with dim(X ) = 0 is scattered. The following theorem states that the converse holds in the class of Lindelöf spaces. This is very useful for showing that certain spaces are zero-dimensional. Theorem 2.4.20 Let X be a non-empty scattered Lindelöf space. Then one has dim(X ) = 0. Proof As X is scattered, it admits a base B consisting of clopen subsets. Consider a finite open cover α = (Ui )i∈I of X . For every x ∈ X , we can find an index i(x) ∈ I such that x ∈ Ui(x) . As B is a base of X , there exists B(x) ∈ B such that x ∈ B(x) ⊂ Ui(x) . The subsets B(x), x ∈ X , form an open cover of X . Since X is Lindelöf, this open cover admits a countable subcover. Therefore there exists a cover β = (Bn )n∈N of X such that β α and Bn ∈ B for all n. Consider the sequence γ = (Cn )n∈N of subsets of X defined by C0 := B0 and Cn := Bn \ (B0 ∪ B1 ∪ · · · ∪ Bn−1 ) , for every integer n ≥ 1. As the subsets Bn are clopen and cover X , it is clear that γ is an open partition of X . On the other hand, we have that γ β α. By applying Proposition 1.1.6, we deduce that D(α) = 0. Thus, we have dim(X ) = supα D(α) = 0. Remark 2.4.21 As mentioned earlier, we shall give in Sect. 5.4 an example of a scattered locally compact Hausdorff space with positive topological dimension. By Corollary 2.3.3, every accessible space X with dim(X ) = 0 is scattered. Combining this result with the previous theorem, we get the following. Corollary 2.4.22 Let X be an accessible Lindelöf space (e.g., a separable metrizable space or a compact Hausdorff space) with X = ∅. Then the following conditions are equivalent:

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(a) dim(X ) = 0; (b) X is scattered.

Example 2.4.23 We deduce from Corollary 2.4.22 and Proposition 2.2.8 that a nonempty subset X ⊂ R satisfies dim(X ) = 0 if and only if X has empty interior in R. This shows in particular that the set R\Q of irrational numbers satisfies dim(R\Q)=0. As an immediate consequence of Corollary 2.4.22, we obtain the following results. Corollary 2.4.24 Let (X i )i∈I be a family of compact Hausdorff spaces with dim(X i ) = 0 for all i ∈ I . Then the product space X := i∈I X i satisfies dim(X ) = 0. Proof By Proposition 2.2.7, the space X is scattered since it is a product of scattered spaces. On the other hand, X is a product of compact Hausdorff spaces and hence also compact and Hausdorff. Corollary 2.4.25 Let (X i )i∈I be a family of non-empty finite discrete spaces. Then the product space X := i∈I X i satisfies dim(X ) = 0. Proof This immediately follows from Corollary 2.4.24 since each X i is a compact Hausdorff space with dim(X i ) = 0. By taking X i = {0, 1} for all i ∈ I in Corollary 2.4.25, we get the following. Corollary 2.4.26 One has dim({0, 1} E ) = 0 for any set E. dim({0, 1}N )

{0, 1}N

Example 2.4.27 We have = 0. As is homeomorphic to the Cantor set K by Proposition 2.1.3, we recover the fact that dim(K ) = 0 (cf. Proposition 2.1.2). Corollary 2.4.28 Let (X i )i∈I be a countable family of separable metrizable spaces such that dim(X i ) = 0 for all i ∈ I . Then the product space X := i∈I X i satisfies dim(X ) = 0. Proof By Proposition 2.2.7, the space X is scattered since it is a product of scattered spaces. On the other hand, X is a product of countably many separable metrizable spaces and hence also separable and metrizable. The following example shows that the product of two zero-dimensional topological spaces may fail to be zero-dimensional. Example 2.4.29 Let X = {x0 , x1 } be a set with cardinality 2. Equip X with the topology for which the open sets are ∅, {x0 } and X . We have that dim(X ) = 0 since the open cover of X reduced to X is finer than any open cover of X . In fact, X is the space described in Example 1.1.11 for n = 0. Consider now the set X × X equipped with the product topology. The open subsets of X × X are ∅ and all the subsets of X × X that contain (x0 , x0 ). Thus, we have that dim(X × X ) = 2 by applying the result in Example 1.1.11 for n = 2. Remark 2.4.30 The topological space X in the previous example is not Hausdorff, not even accessible since {x0 } is not closed in X . In Sect. 5.5, we shall give an example of a normal Hausdorff space X such that dim(X ) = 0 and dim(X × X ) = 0.

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2.5 Totally Disconnected Spaces Let X be a topological space. Recall that the connected component of a point x ∈ X is the union of all the connected subsets of X containing x. The connected components of the points of X form a partition of X . Moreover, every connected component is connected and closed in X . Definition 2.5.1 We say that a topological space X is totally disconnected if the connected component of every point x ∈ X is the singleton set reduced to the point x. In other words, a topological space X is totally disconnected if and only if the only non-empty connected subsets of X are the subsets that are reduced to a single point. Example 2.5.2 Every discrete space is totally disconnected. Example 2.5.3 The only connected subsets of R are the intervals. It follows that a subset X ⊂ R is totally disconnected if and only if X has empty interior. Proposition 2.5.4 Every subset of a totally disconnected space is itself totally disconnected. Proof This immediately follows from the observation that if Y is a subset of a topological space X and y ∈ Y then the connected component of y in Y is contained in the connected component of y in X . Proposition 2.5.5 Every product of totally disconnected spaces is itself totally disconnected. Proof Let (X i )i∈I be a family of totally disconnected spaces and consider their direct product X := i∈I X i . Let C be a non-empty connected subset of X . As the continuous image of a connected space is itself connected, the projection of C on each X i is connected and hence reduced to a single point since X i is totally disconnected. This implies that C itself is reduced to a single point. Proposition 2.5.6 Every totally disconnected space is accessible. Proof In a topological space, every connected component is closed. Consequently, if the topological space X is totally disconnected then {x} is closed in X for all x ∈ X. The following example shows that a totally disconnected space may fail to be Hausdorff. Example 2.5.7 Let X be an infinite set. Let us fix two distinct points a, b ∈ X and let Y := X \{a, b}. Let T denote the set consisting of all U ⊂ X satisfying one of the following two conditions:

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(1) U ⊂ Y ; (2) U = U1 ∪ U2 , where U1 is a non-empty subset of {a, b} and U2 ⊂ Y is such that Y \U2 is a finite set. It is straightforward to verify that T is the set of open sets for a topology on X . Let us equip X with this topology. Suppose that A ⊂ X has more than one point. If we can find a point y0 ∈ A ∩ Y , then the singleton set {y0 } is clopen in A. Otherwise, we have that A = {a, b} and then {a} is clopen in A. It follows that A is not connected. Thus, the space X is totally disconnected. However, X is not Hausdorff since every open neighborhood of a meets every open neighborhood of b.

2.6 Totally Separated Spaces In this section, we introduce the class of totally separated spaces. We prove that every totally separated space is totally disconnected and that every scattered accessible space is totally separated. Let X be a topological space. The quasi-component of a point x ∈ X is the intersection of all clopen neighborhoods of x. Note that the quasi-component of every point x ∈ X is a closed subset of X containing x. Definition 2.6.1 We say that a topological space X is totally separated if the quasicomponent of every point x ∈ X is the singleton set reduced to the point x. Remark 2.6.2 A topological space X is totally separated if and only if it satisfies the following condition: for every pair of distinct points x and y in X , there exists a partition of X into two open subsets U and V such that x ∈ U and y ∈ V . Proposition 2.6.3 Every totally separated space is Hausdorff. Proof This immediately follows from the preceding remark.

Proposition 2.6.4 Let X be a topological space and x a point in X . Then the connected component of x is contained in the quasi-component of x. Proof Denote by C x the connected component of x and by Q x its quasi-component. Consider a clopen neighborhood V of x in X . Then C x ∩ V is a clopen subset of C x that is not empty since it contains x. By connectedness of C x , we deduce that C x ∩ V = C x , that is, C x ⊂ V . It follows that C x ⊂ Q x . Corollary 2.6.5 Every totally separated space is totally disconnected.

A totally disconnected space is not necessarily totally separated. Indeed, we have described in Example 2.5.7 a totally disconnected space that is not Hausdorff. Such a space is not totally separated since, by Proposition 2.6.3, every totally separated space is Hausdorff. In Sect. 5.2, we shall give an example of a totally disconnected separable metrizable space that is not totally separated.

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Proposition 2.6.6 Every scattered accessible space is totally separated and hence totally disconnected. Proof Let X be a scattered accessible space. Let B be a base of X consisting of clopen subsets of X . Consider a point x in X . As B is a base of X , the set Bx consisting of all elements of B containing x is a neighborhood base of x. The intersection of all the neighborhoods of x is reduced to the point x since X is accessible. This implies that the intersection of the elements of Bx is also reduced to x. Consequently, the quasi-component of x is the singleton set {x}. This shows that X is totally separated. The accessibility hypothesis in Proposition 2.6.6 cannot be removed. Indeed, a set having more than one point equipped with its trivial topology is scattered but not totally separated (not even totally disconnected). Let us note also that the converse of Proposition 2.6.6 is false. Indeed, we will give in Sect. 5.1 an example of a separable metrizable space that is totally separated but not scattered. However, as we shall see, the converse of Proposition 2.6.6 becomes true if we restrict ourselves to locally compact Hausdorff spaces. Let us first establish the following result. Lemma 2.6.7 Let X be a compact Hausdorff space. Let x be a point in X . Then the connected component of x coincides with its quasi-component. Proof Denote by C x the connected component of x and by Q x its quasi-component. We have that C x ⊂ Q x by Proposition 2.6.4. Thus, it suffices to prove that Q x is connected. Let A and B be disjoint closed subsets of Q x such that A ∪ B = Q x . We can assume that x ∈ A. As Q x is closed in X , the sets A and B are closed in X . On the other hand, since X is a compact Hausdorff space, it is normal by Proposition 1.5.4. Consequently, there exist disjoint open subsets V and W of X such that A ⊂ V and B ⊂ W . Denote by E the set consisting of all clopen neighborhoods of x in X . We have that U = Q x ⊂ V ∪ W. U ∈E

Therefore, the open subsets X \U , U ∈ E, cover X \(V ∪ W ). As X \(V ∪ W ) is compact, there exists a finite sequence U1 , . . . , Un of elements of E such that X \(V ∪ W ) ⊂ (X \U1 ) ∪ · · · ∪ (X \Un ). By setting := U1 ∩ · · · ∩ Un , this amounts to saying that ⊂ V ∪ W . As V and W are disjoint, we deduce that ∩ V = \W . Consequently, the set ∩ V is a clopen neighborhood of x in X . It follows that Q x ⊂ ∩ V . Therefore we have that Q x = A. This shows that Q x is connected.

2.6 Totally Separated Spaces

43

Proposition 2.6.8 Let X be a locally compact Hausdorff space. Then the following conditions are equivalent: (a) X is scattered; (b) X is totally separated; (c) X is totally disconnected. Proof The fact that (a) implies (b) follows from Proposition 2.6.6. On the other hand, Corollary 2.6.5 shows that (b) implies (c). Suppose that X is totally disconnected. Let x be a point in X and let V be a neighborhood of x. As X is locally compact, there exists a compact neighborhood W of x such that W ⊂ V . Denote by U the interior of W in X and by E the set consisting of all clopen neighborhoods of x in W . As W is totally disconnected by

Proposition 2.5.4, it follows from Lemma 2.6.7 that {x} = F∈E F. This implies that the family α := {U } ∪ {W \F | F ∈ E} is an open cover of W . Since W is compact, α admits a finite subcover. This means that there exists a finite sequence F1 , . . . , Fn ∈ E such that the set A := F1 ∩· · ·∩ Fn satisfies A ⊂ U . Each Fi , 1 ≤ i ≤ n, is closed in W and hence in X since W is closed in X . On the other hand, A is open in U and hence open in X . It follows that A is clopen in X . As x ∈ A ⊂ V , we deduce that the neighborhoods of x that are clopen in X form a neighborhood base of x. This shows that X is scattered. Thus, (c) implies (a).

2.7 Zero-Dimensional Compact Hausdorff Spaces By combining results obtained in the previous sections, we get the following characterizations of zero-dimensional compact Hausdorff spaces. Theorem 2.7.1 Let X be a non-empty topological space. Then the following conditions are equivalent: (a) (b) (c) (d) (e)

X is a compact Hausdorff space with dim(X ) = 0; X is a scattered compact Hausdorff space; X is a totally separated compact Hausdorff space; X is a totally disconnected compact Hausdorff space; there exists a set E such that X is homeomorphic to a closed subset of the product space {0, 1} E .

Proof Conditions (a) and (b) are equivalent by virtue of Corollary 2.4.22. On the other hand, conditions (b), (c) and (d) are equivalent by Proposition 2.6.8. Finally, the equivalence of (b) and (e) is an immediate consequence of Proposition 2.2.9 since the product space {0, 1} E is a compact Hausdorff space for any set E by Tychonoff’s theorem.

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2.8 Zero-Dimensional Separable Metrizable Spaces We also get the following characterizations of zero-dimensional separable metrizable spaces. Theorem 2.8.1 Let X be a non-empty topological space. Then the following conditions are equivalent: (a) X is a separable metrizable space with dim(X ) = 0; (b) X is a scattered separable metrizable space; (c) X is a separable metrizable space that admits a countable base consisting of clopen subsets; (d) X is homeomorphic to a subset of {0, 1}N ; (e) X is homeomorphic to a subset of the Cantor set. Proof Conditions (a) and (b) are equivalent by Corollary 2.4.22. Suppose that X is a scattered separable metric space. Let B be a base of X consisting of clopen subsets. As X is separable, we can find a countable dense subset Y ⊂ X . Let us choose, for each y ∈ Y and each integer n ≥ 1, a neighborhood B y,n ∈ B of y contained in the open ball of radius 1/n centered at y. Then the subsets B y,n form a countable base of X . This shows that (b) implies (c). Let us now show that (c) implies (d) (cf. the proof of Proposition 2.2.9). Suppose that X is a separable metric space and that (Bn )n∈N is a base of X consisting of clopen subsets. Let χn : X → {0, 1} denote the characteristic map of Bn . Consider the map ϕ : X → {0, 1}N defined by ϕ(x) = (χn (x))n∈N for all x ∈ X . As Bn is clopen, the map χn is continuous for every n ∈ N. This implies that ϕ is continuous. As X is Hausdorff, the injectivity of ϕ follows from the fact that the subsets Bn , n ∈ N, form a base of X . We have that ϕ(Bn ) = ϕ(X ) ∩ πn−1 (1), where πn : {0, 1}N → {0, 1} is the projection onto the n-factor of {0, 1}N . This shows that ϕ(Bn ) is open in ϕ(X ). As the subsets Bn form a base of X , we deduce that the image by ϕ of any open subset of X is open in ϕ(X ). Consequently, ϕ induces a homeomorphism from X onto ϕ(X ). This shows that X satisfies (d). To complete the proof, it suffices to observe that (d) implies (b) by Proposition 2.2.9 and that (d) and (e) are equivalent since the space {0, 1}N is homeomorphic to the Cantor set by Proposition 2.1.3. Remark 2.8.2 As already mentioned above, we will give in Sect. 5.1 an example of a separable metrizable space that is totally separated (and hence totally disconnected) but not scattered.

2.9 Zero-Dimensional Compact Metrizable Spaces Every compact metrizable space is both Hausdorff and separable. By combining Theorems 2.7.1 and 2.8.1, we obtain the following statement (Table 2.1).

2.9 Zero-Dimensional Compact Metrizable Spaces

45

Table 2.1 Summary Table (X non-empty) NOTES

dim(X) = 0 ⇒ scattered ⇒ totally sepa- ⇒ totally disrated connected Hausdorff spaces dim(X) = 0 ⇔ scattered ⇒ totally sepa- ⇒ totally disrated connected separable metrizable spaces dim(X) = 0 ⇒ scattered ⇔ totally sepa- ⇔ totally disrated connected locally compact Hausdorff spaces dim(X) = 0 ⇔ scattered ⇔ totally sepa- ⇔ totally disrated connected compact Hausdorff spaces Theorem 2.9.1 Let X be a non-empty topological space. Then the following conditions are equivalent: X is a compact metrizable space with dim(X ) = 0; X is a scattered compact metrizable space; X is a totally separated compact metrizable space; X is a totally disconnected compact metrizable space; X is a compact metrizable space that admits a countable base consisting of clopen subsets; (f) X is homeomorphic to a closed subset of {0, 1}N ; (g) X is homeomorphic to a closed subset of the Cantor set.

(a) (b) (c) (d) (e)

Notes The terminology used in this chapter follows that of Bourbaki [18]. However, the terms “scattered”, “totally disconnected”, and “totally separated” have sometimes different meanings in the literature. For example, spaces that are called “scattered” in the present book are called “zero-dimensional” in [102], while a “scattered” space in [102] is a topological space in which every non-empty subset admits an isolated point.

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2 Zero-Dimensional Spaces

The Cantor ternary set was described by Cantor in [21, note 11 p. 46]. It can be shown that every totally disconnected compact metrizable space that is perfect is homeomorphic to the Cantor set (see for example [48, Corollary 2–98]). A non-empty topological space X is scattered if and only if ind(X ) = 0 (see the Notes on Chap. 1, p.19, for the definition of the small inductive dimension ind(X )). The question of the existence of scattered metrizable spaces with positive topological dimension remained open for many years (cf. [18, note 1 p. IX.119]). An affirmative answer to this question was finally given by Roy [96, 98] who constructed a scattered metrizable space X with dim(X ) = 1. The notion of a totally disconnected space and that of a totally separated space were respectively introduced by Hausdorff [47] and by Sierpinski [99]. In [99], Sierpinski described a totally disconnected subset of R2 that is not totally separated and a totally separated subset of R2 with positive topological dimension.

Exercises 2.1 2.2 2.3 2.4

Does the real number 1/π belong to the Cantor set? Show that the Cantor set has Lebesgue measure 0. Show that every countable product of Cantor spaces is a Cantor space. Let H denote the Hilbert space of square-summable real sequences (u n )n≥1 . Show that the subset X ⊂ H consisting of all sequences (u n )n≥1 such that |u n | ≤ 1/n for all n ≥ 1 is homeomorphic to the Hilbert cube [0, 1]N . 2.5 Let G be a group. Let B denote the set of all left cosets of subgroups of finite index of G, i.e., the subsets of the form g H , where g ∈ G and H ⊂ G is a subgroup with [G : H ] < ∞. (a) Show that there is a unique topology on G admitting B as a base. This topology is called the profinite topology on G. (b) Show that the profinite topology on G is scattered. (c) Show that the profinite topology on G is discrete if and only if G is finite. (d) Show that the profinite topology on the additive group Q of rational numbers is the trivial topology. (e) Show that the profinite topology on G is Hausdorff if and only if G is residually finite. (Recall that the group G is called residually finite if the intersection of all its subgroups of finite index is reduced to the identity element.) 2.6 (Furstenberg’s topological proof of the infinitude of primes [38]). Let Z denote the group of integers equipped with its profinite topology (see Exercise 2.5). (a) Show that nZ is a closed subset of Z for every n ∈ Z. (b) Show that every non-empty open subset of Z is infinite.

Exercises

47

(c) Let P := {2, 3, 5, 7, 11, . . . } denote the set of prime numbers. Use the results obtained in (a) and (b) to recover Euclid’s theorem that P is infinite. Hint: observe that p∈P pZ = Z\{−1, 1} is not closed in Z. 2.7 Let f : X → Y be a continuous map from a Lindelöf space X into a topological space Y . Show that f (X ) is a Lindelöf space. 2.8 Show that every locally compact Lindelöf space is σ -compact. 2.9 Let X be an uncountable set equipped with its cofinite topology. Show that X is not first-countable. 2.10 Show that every open subset of a separable space is separable. 2.11 Show that every subspace of a separable metrizable space is separable. 2.12 Show that the set consisting of all isolated points of a separable space is countable. 2.13 Show that every countable product of separable spaces is separable. 2.14 Let (X, d) be a separable metric space. Consider the Banach space ∞ (R) consisting of all bounded sequences of real numbers u = (u n )n∈N with the supremum norm u = supn∈N |u n |. Fix a point x0 ∈ X and a sequence (an )n∈N of points of X such that the set {an | n ∈ N} is dense in X . Show that the sequence (d(x, an ) − d(x0 , an ))n∈N is in ∞ (R) for every x ∈ X and that the map ϕ : X → ∞ (R) defined by ϕ(x) = (d(x, an ) − d(x0 , an ))n∈N is an isometric embedding. 2.15 Show that the Banach space ∞ (R) is not separable. 2.16 Show that every second-countable scattered accessible space is homeomorphic to a subset of the Cantor set. 2.17 A metric space (X, d) is called an ultrametric space if one has d(x, y) ≤ max(d(x, z), d(y, z)) for all x, y, z ∈ X . Let (X, d) be a non-empty ultrametric space. (a) Let A be a closed subset of X and ρ > 0. Show that the set consisting of all x ∈ X such that dist(x, A) = ρ is a clopen subset of X . (b) Let A and B be disjoint closed subsets of X . Show that the set consisting of all x ∈ X such that dist(x, A) ≤ dist(x, B) is a clopen subset of X . (c) Show that dim(X ) = 0. (d) Show that the metric completion (X , d ) of (X, d) is also an ultrametric space. 2.18 Let p be a prime integer. Every non-zero rational number q ∈ Q\{0} can be a written in the form q = p n , where n ∈ Z and a, b ∈ Z\ pZ are integers b not divisible by p. The integer v p (q) := n ∈ Z is well defined and called the p-valuation of q. Define the map d : Q × Q → R by d(x, y) :=

p −v p (x−y) 0

if x = y if x = y

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2 Zero-Dimensional Spaces

for all x, y ∈ Q. (a) Show that (Q, d) is an ultrametric space. (b) Show that the metric completion Q p of (Q, d) satisfies dim(Q p ) = 0. (The set Q p is the set of p-adic numbers.) 2.19 Show that every totally disconnected topological space that is locally connected is discrete. (Recall that a topological space X is called locally connected if every point x ∈ X admits a neighborhood base consisting of connected subsets.) 2.20 Let X be a non-empty subset of R. Show that one has dim(X ) = 0 if and only if X is totally disconnected. 2.21 Let X be the topological space described in Example 2.5.7. (a) Show that X is compact. (b) Show that X is not normal. (c) Show that dim(X ) = 1. 2.22 A topological space X is called extremally disconnected if the closure of any open subset of X is open in X . (a) Show that if a set X is equipped with its trivial (resp. discrete) topology then X is extremally disconnected. (b) Show that every extremally disconnected Hausdorff space is totally separated. (c) Show that every extremally disconnected metrizable space is discrete.

Chapter 3

Topological Dimension of Polyhedra

In this chapter, we introduce the notion of a simplicial complex of Rn and that of a polyhedron. A simplicial complex is a finite assembly of simplices and a polyhedron is a topological space that is homeomorphic to some simplicial complex. The main results of this chapter is that the unit cube in Rn has topological dimension n for any integer n ≥ 0 (Theorem 3.5.4). This is used to shows that the topological dimension of the support of a simplicial complex is equal to its combinatorial dimension (Corollary 3.5.5). We also deduce that the topological dimension of the Euclidean space Rn is n (Corollary 3.5.7).

3.1 Simplices of Rn Let us start by briefly reviewing some elementary facts about affine subspaces of Rn . This material is standard and proofs are omitted. Let n ≥ 0 be an integer. A subset A ⊂ Rn is called an affine subspace of Rn if either A = ∅ or there exist a linear subspace V ⊂ Rn and a point p ∈ Rn such that A = p + V := { p + v | v ∈ V }. In this case, we have that V = {b − a | a, b ∈ A}. It follows in particular that the linear subspace V ⊂ Rn depends only on A. One says that V is the direction of the affine subspace A. The dimension dimR (V ) of V as a vector space over R is called the affine dimension of A and is denoted dimaff (A). By convention, the affine dimension of the empty set ∅ is dimaff (∅) = −1. An affine hyperplane of Rn is an affine subspace of affine dimension n − 1. The affine subspace of Rn generated by k + 1 points p0 , p1 , . . . , pk ∈ Rn is the smallest affine subspace A ⊂ Rn such that pi ∈ A for all 0 ≤ i ≤ k. One has

© Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_3

49

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3 Topological Dimension of Polyhedra

A={

k

λi pi | λi ∈ R (0 ≤ i ≤ k) and

k

i=0

λi = 1}.

i=0

The points p0 , p1 , . . . , pk ∈ Rn are said to be affinely independent if the affine subspace of Rn they generate has affine dimension k. This amounts to saying that the k vectors p1 − p0 , . . . , pk − p0 are linearly independent in Rn . Let p0 , p1 , . . . , pk be affinely independent points in Rn . Consider the convex hull := [ p0 , p1 , . . . , pk ] ⊂ Rn of the set { p0 , p1 , . . . , pk }. We have that = [ p0 , p1 , . . . , pk ] = {

k

λi pi | λi ≥ 0 (0 ≤ i ≤ k) and

i=0

k

λi = 1}.

i=0

The extremal points of are the points p0 , p1 , . . . , pk (recall that a point x of a convex subset K ⊂ Rn is called extremal if the set K \{x} is convex). One says that is the simplex whose vertex set is { p0 , p1 , . . . , pk }. Observe that is compact since it is a closed and bounded subset of Rn . The integer k (i.e., the number of vertices of minus 1) is called the combinatorial dimension of the simplex and is denoted dimcomb (). Note that dimcomb () = dima f f (A), where A is the affine subspace of Rn generated by the points p0 , p1 , . . . , pk . When is a simplex with combinatorial dimension k, one also says that is a k-simplex. Thus, a 0-simplex is a subset of Rn reduced to a single point, a 1-simplex is a line segment joining two distinct points of Rn , a 2-simplex is a triangle, a 3-simplex is a tetrahedron, etc. (see Fig. 3.1). The empty set is a simplex whose vertex set is empty and whose combinatorial dimension is −1. Suppose that p0 , p1 , . . . , pk are affinely independent points in Rn and let denote the simplex whose vertex set is { p0 , p1 , . . . , pk }. Consider a subset I ⊂ {0, 1, . . . , k}. Then the points pi , where i ∈ I , are affinely independent. They are the vertices of a simplex ⊂ whose combinatorial dimension is dimcomb ( ) = −1 + #I . One says that the simplex is a face of . The number of faces of is equal to the number of subsets of {0, 1, . . . , k}, i.e., 2k+1 . Two particular faces of are the empty set ∅ and the simplex itself. They are obtained by taking I = ∅ and I = {0, 1, . . . , k} respectively. A face of is called proper if ∅ = = .

k= 0

Fig. 3.1 Some k-simplices

k= 1

k= 2

k= 3

3.1 Simplices of Rn

51

◦

The set ⊂ defined by ◦

:= {

k

λi pi | λi > 0 (0 ≤ i ≤ k) and

i=0

k

λi = 1}

i=0 ◦

is called the open simplex with vertices p0 , p1 , . . . , pk . Observe that is the interior ◦

of in the affine subspace A of Rn generated by the vertices of . In particular, ◦

is open in A. Note also that \ is the union of the proper faces of . We have that ◦

◦

= for k ≥ 1. For k = 0, we have that = = { p0 }.

3.2 Simplicial Complexes of Rn Definition 3.2.1 A simplicial complex of Rn is a finite set C of simplices of Rn satisfying the following conditions: SC1 if ∈ C, then every face of belongs to C; SC2 if 1 ∈ C and 2 ∈ C, then 1 ∩ 2 is a face of both 1 and 2 (Fig. 3.2). Let C be a simplicial complex of Rn . A point p ∈ Rn is a vertex of C if there exists a simplex ∈ C such that p is one of the vertices of . As C contains only finitely many simplices, the set of vertices of C is finite. It follows from conditions (SC1) and (SC2) of Definition 3.2.1 that if p is a vertex of C and is a simplex of C such that p ∈ , then p is a vertex of . The combinatorial dimension dimcomb (C) of the simplicial complex C is the maximal combinatorial dimension of the simplices of C. Note that we always have dimcomb (C) ≤ n. The set |C| ⊂ Rn defined by |C| :=

∈C

is called the support of the simplicial complex C.

Fig. 3.2 A simplicial complex of R2

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3 Topological Dimension of Polyhedra

Definition 3.2.2 A topological space X is called a polyhedron if there exist an integer n ≥ 0 and a simplicial complex C of Rn such that X is homeomorphic to the support |C| of C. Note that every polyhedron is compact and metrizable. ◦

Proposition 3.2.3 Let C be a simplicial complex of Rn . Then the open simplices , where ∈ C, form a partition of the support |C| of C. Proof Let x ∈ |C|. Choose a simplex 0 ∈ C containing x. Let k := dimcomb (0 ) and suppose that p0 , p1 , . . . , pk are the vertices of 0 . As x ∈ 0 , there exist real numbers λi ≥ 0, 0 ≤ i ≤ k, such that x=

k

k

λi pi and

i=0

λi = 1.

i=0

Denote by I the set consisting of all i ∈ {0, 1, . . . , k} such that λi > 0. The simplex whose vertices are the points pi , with i ∈ I , belongs to C by (SC1) since it is a face of 0 . On the other hand, we have that λi pi and λi = 1. x= i∈I

i∈I ◦

As λi > 0 for all i ∈ I , we deduce that x ∈ .

◦

Suppose now that there is another simplex ∈ C such that x ∈ . By (SC2), the ◦

set ∩ is a face of both and . The simplex ∩ meets since x ∈ ∩ . ◦

As any proper face of is contained in \, we deduce that ∩ = . Similarly, we get ∩ = . It follows that = . Remark 3.2.4 Note that if C is a simplicial complex of Rn and is a simplex ◦

belonging to C, then the open simplex is not necessarily open in |C|.

3.3 Open Stars In this section, given a simplicial complex C of Rn and a vertex p of C, we define the open star of C at p. We prove (see Proposition 3.3.2) that the open stars of C form a finite open cover of the support |C| of C and that the order of this cover is equal to the combinatorial dimension of C. Consider a simplicial complex C of Rn . Let p be one of the vertices of C. Denote by FC ( p) the union of all the simplices of C that do not contain p. As every simplex

3.3 Open Stars

53

is compact and C contains only finitely many simplices, the set FC ( p) is compact and hence closed in |C|. It follows that the set StC ( p) := |C|\FC ( p) is an open neighborhood of p in |C|. We call StC ( p) the open star of the simplicial complex C at the vertex p. Proposition 3.3.1 One has StC ( p) =

◦

.

(3.3.1)

p∈∈C

Proof Denote by E the right-hand side of (3.3.1). Let x ∈ StC ( p). By Proposition 3.2.3, there exists a simplex ∈ C such that ◦

x ∈ . As x ∈ / FC ( p), we have that p ∈ and hence x ∈ E. This shows that StC ( p) ⊂ E. Conversely, suppose now that x ∈ E. This means that there exists ∈ C such ◦

that x ∈ and p ∈ . Consider a simplex 1 ∈ C such that x ∈ 1 . Then there is ◦

a face 2 of 1 such that x ∈ 2 . By applying Proposition 3.2.3, we get 2 = . / FC ( p). This shows This implies p ∈ 2 and hence p ∈ 1 . We conclude that x ∈ that E ⊂ StC ( p). Proposition 3.3.2 Let C be a simplicial complex of Rn . Let V denote the set of vertices of C. Then the family α := (StC ( p)) p∈V is a finite open cover of the support |C| of C. Moreover, the order of α is equal to the combinatorial dimension of the simplicial complex C. Proof The set V is finite since C contains only finitely many simplices. On the other hand, we have seen above that StC ( p) is open in |C| for all p ∈ V . Therefore α is a finite family of open subsets of |C|. Let x be a point in |C|. By Proposition 3.2.3, there is a simplex ∈ C such that ◦

x ∈ . Let p be one of the vertices of . It follows from Proposition 3.3.1 that x ∈ StC ( p). Consequently, α covers |C|. Let k := ord(α). Then there exist distinct vertices p0 , p1 , . . . , pk of C such that k

StC ( pi ) = ∅.

i=0

k Let x ∈ i=0 StC ( pi ) and choose a simplex ∈ C such that x ∈ . As x ∈ StC ( pi ), we have that pi ∈ for all i = 0, 1, . . . , k. Consequently, the points p0 , p1 , . . . , pk are vertices of . This shows that ord(α) = k ≤ dimcomb (C).

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Consider now a non-empty simplex ∈ C. Let p0 , p1 , . . . , pk denote the vertices ◦ k of . As i=0 StC ( pi ) ⊃ = ∅, we have that ord(α) ≥ k. It follows that ord(α) ≥ dimcomb (C).

3.4 Barycentric Subdivisions Let ⊂ Rn be a non-empty k-simplex whose vertices are p0 , p1 , . . . , pk . The ◦

barycenter of the simplex is the point γ ∈ defined by γ :=

1 ( p0 + p1 + · · · + pk ). k+1

Given two simplices and of Rn , we write < if is a proper face of Note that the relation < is transitive. A chain of simplices of Rn is a (possibly empty) finite sequence

.

π = (0 , 1 , . . . , r ) of non-empty simplices of Rn such that 0 < 1 < · · · < r . The integer r ∈ {−1, 0, . . . , n} is called the length of the chain π. Let π = (0 , 1 , . . . , r ) be a chain of simplices of Rn . Denote by γi the barycenter of the simplex i (0 ≤ i ≤ r ). Clearly the points γ0 , γ1 , . . . , γr −1 belong to the simplex r −1 and γr is not contained in the affine subspace of Rn generated by r −1 . By induction on r , we deduce that the points γ0 , γ1 , . . . , γr are affinely independent in Rn . Consequently, these points are the vertices of an r -simplex (π) = [γ0 , γ1 , . . . , γr ] ⊂ Rn . The simplex (π) is called the simplex associated with the chain π. We have that (π) ⊂ r since all the simplices of (π) are contained in r . Let C be a simplicial complex of Rn . We say that a chain π = (0 , 1 , . . . , r ) of simplices of Rn is a C-chain if i ∈ C for all i ∈ {0, 1, . . . , r }. We denote by (C) the set consisting of all C-chains. ◦

Lemma 3.4.1 Let C be a simplicial complex of Rn . Then the open simplices (π), π ∈ (C), form a partition of the support |C| of C.

3.4 Barycentric Subdivisions

55

Proof Every chain π = (0 , 1 , . . . , r ) ∈ (C) satisfies (π) ⊂ |C| since (π) ⊂ r . Let x ∈ |C|. By Proposition 3.2.3, there exists a unique simplex ∈ C such that ◦

x ∈ . Let S denote the set of vertices of . There is a unique family of positive real numbers (λ p ) p∈S such that x=

λ p p and

p∈S

λ p = 1.

(3.4.1)

p∈S

Let us write {λ p | p ∈ S} = {α0 , α1 , . . . , αr } with α0 > α1 > · · · > αr . For each i ∈ {0, 1, . . . , r }, denote by Ti the set consisting of all vertices p ∈ S such that λ p = αi . The sets T0 , T1 , . . . , Tr are all non-empty and form a partition of S. By rearranging the terms of the right-hand side of (3.4.1), we get ⎞ ⎛ r x= (3.4.2) αi ⎝ p⎠ . i=0

p∈Ti

Let us set, for each i ∈ {0, 1, . . . , r }, Si := T0 ∪ T1 ∪ · · · ∪ Ti and denote by i the simplex whose vertex set is Si . We clearly have 0 < 1 < · · · < r = . It follows that π := (0 , 1 , . . . , r ) is a C-chain. After summing up by parts, equality (3.4.2) gives us ⎛ ⎞ r −1 x= (αi − αi+1 ) ⎝ p ⎠ + αr p. i=0

p∈Si

p∈Sr

Denoting by γi the barycenter of i (0 ≤ i ≤ r ), we obtain x=

r i=0

μi γi ,

56

3 Topological Dimension of Polyhedra

where μi := (αi − αi+1 )#Si for 0 ≤ i ≤ r − 1 and μr := αr #Sr . It is easy to verify ◦

that μi > 0 for all i and ri=1 μi = 1. Therefore we have that x ∈ (π). Suppose now that there exists another C-chain π = (0 , 1 , . . . , r ) ◦

such that x ∈ (π ). Then, denoting by γi the barycenter of i , there exist real numbers μi > 0 (0 ≤ i ≤ r ) such that

x=

r

μi γi

and

i=0

r

μi = 1.

i=0

Let Si denote the set of vertices of i (0 ≤ i ≤ r ). Let us set T0 := S0 and for all i ∈ {1, . . . , r }. We have that Ti := Si \Si−1

x=

r

⎛ αi ⎝

i=0

⎞

p⎠ ,

p∈Ti

where αi :=

μi+1 μr μi + + · · · + (0 ≤ i ≤ r ). #Si #Si+1 #Sr

The sets T0 , T1 , . . . , Tk are non-empty and form a partition of Sr . On the other hand, we have that

α0

>

α1

> ··· >

αr

> 0 and

r

αi #Ti

i=0

=

r

μi = 1.

i=0

By using Proposition 3.2.3, we deduce that r = r and Ti = Ti for all i ∈ {0, 1, . . . , r }. It follows that Si = Si for all i ∈ {0, 1, . . . , r }. This shows that π = π . Proposition 3.4.2 Let C be a simplicial complex of Rn . Then the set C := {(π) | π ∈ (C)}

(3.4.3)

is a simplicial complex of Rn . Moreover, the simplicial complexes C and C have the same combinatorial dimension and the same support.

3.4 Barycentric Subdivisions

57

Proof Let π = (0 , 1 , . . . , r ) be a C-chain. Denote by γi the barycenter of i be a face of (π). Denote by s the combinatorial dimension (0 ≤ i ≤ r ). Let , where 0 ≤ i 0 < i 1 < and suppose that γi0 , γi1 , . . . , γis are the vertices of of · · · < i s ≤ r . Clearly π := (i0 , i1 , . . . , is ) is a C-chain and we have that = ( π ) ∈ C . This shows that C satisfies condition (SC1) of Definition 3.2.1. Let us show now that C satisfies condition (SC2). Let π = (0 , 1 , . . . , r ) and π = (0 , 1 , . . . , r ) be two C-chains. The simplices appearing in both π and π form a C-chain π = (0 , 1 , . . . , r ). We claim that (π) ∩ (π ) = (π ).

(3.4.4)

The inclusion (π ) ⊂ (π) ∩ (π ) is obvious since the simplex (π ) is a face of both (π) and (π ). Let x ∈ (π) ∩ (π ). As x ∈ (π), we can extract a ◦ of π such chain π of π such that x ∈ ( π ). Similarly, we can extract a chain π ◦

. It follows that the chain that x ∈ ( π ). Lemma 3.4.1 implies that π =π π can be extracted from the chain π and hence that x ∈ (π ). This completes the proof of (3.4.4). This shows that C satisfies (SC2). The inclusion |C | ⊂ |C| is straightforward. The inclusion |C| ⊂ |C | follows from Lemma 3.4.1. It follows that |C| = |C |. Let m := dimcomb (C) and m := dimcomb (C ). If π = (0 , 1 , . . . , r ) is a C-chain, then the length r of π is clearly less than or equal to the combinatorial dimension of r . It follows that m ≤ m. Let be a m-simplex in C. If p0 , p1 , . . . , pm are the vertices of , then ([ p0 ], [ p0 , p1 ], . . . , [ p0 , p1 , . . . , pm ]) is a C-chain with length m. It follows that m ≤ m . This shows that the simplicial complexes C and C have the same combinatorial dimension. The simplicial complex C is called the barycentric subdivision of C. Suppose now that Rn is equipped with a metric d. If C is a simplicial complex of Rn , we define the mesh by mesh(C) := max diam(). ∈C

Proposition 3.4.3 Let V denote the vertex set of C and α := (StC ( p)) p∈V the finite open cover of |C| consisting of the open stars of C. Then one has mesh(α) ≤ 2 mesh(C). Proof Let p ∈ V . Suppose that x and y are two points in StC ( p). Then there exist simplices 1 and 2 in C admitting p as a vertex and containing x and y respectively. By applying the triangle inequality, we obtain

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3 Topological Dimension of Polyhedra

d(x, y) ≤ d(x, p) + d( p, y) ≤ diam(1 ) + diam(2 ) ≤ 2 mesh(C). This implies diam(StC ( p)) ≤ 2 mesh(C) for all p ∈ V . It follows that mesh(α) ≤ 2 mesh(C). We assume from now on that d is the metric associated with a norm · on Rn , that is, d(x, y) = x − y for all x, y ∈ Rn . Lemma 3.4.4 Let ⊂ Rn be a k-simplex whose vertices are the points p0 , p1 , . . . , pk . Then one has (3.4.5) diam() = max pi − p j . i, j

Proof Let δ denote the right-hand side of (3.4.5). We have that δ ≤ diam() since pi , p j ∈ for all i, j. Consider the closed d-ball Bi of radius δ centered at pi . Then we have that p j ∈ Bi for all j. As Bi is convex, it follows that ⊂ Bi . We deduce that if x ∈ then d(x, pi ) ≤ δ for all i. Consequently, the simplex is contained in the closed dball of radius δ centered at x. It follows that d(x, y) ≤ δ for all x, y ∈ , that is, diam() ≤ δ. Proposition 3.4.5 Let C be a simplicial complex of Rn with dimcomb (C) = m. Let C be the barycentric subdivision of C. Then one has mesh(C ) ≤

m mesh(C). m+1

Proof Let be a simplex in C . Consider two distinct vertices u and v of . After possibly exchanging u and v, we can find two simplices 1 < 2 of C whose barycenters are u and v respectively. Let r := dimcomb (1 ) and s := dimcomb (2 ). Let p0 , p1 , . . . , ps be the vertices of 2 numbered in such a way that 1 = [ p0 , p1 , . . . , pr ]. Let w denote the barycenter of the simplex [ pr +1 , . . . , ps ]. Then, we have 1 ( p0 + p1 + · · · + ps ) s+1 1 = ((r + 1)u + (s − r )w), s+1

v=

which gives us u−v =

s −r (u − w). s+1

We deduce that u − v =

s −r u − w . s+1

3.4 Barycentric Subdivisions

59

As the points u and w belong to 2 , this equality implies s −r diam(2 ). s+1

u − v ≤ Since diam(2 ) ≤ mesh(C) and

s m s −r ≤ ≤ , s+1 s+1 m+1 it follows that m mesh(C). m+1

u − v ≤ By applying Lemma 3.4.4, we obtain diam( ) ≤

m mesh(C). m+1

mesh(C ) ≤

m mesh(C). m+1

This shows that

Let C be a simplicial complex of Rn . One defines by induction the N -th barycentric subdivision C (N ) of C by setting C (0) = C and C (i+1) = (C (i) ) for any integer i ≥ 0. Proposition 3.4.6 Let C be a simplicial complex of Rn . Then one has lim mesh(C (N ) ) = 0.

N →∞

Proof The simplicial complexes C (N ) , N ≥ 0, have the same combinatorial dimension by Proposition 3.4.2. Therefore, by applying Proposition 3.4.5, we get mesh(C (N ) ) ≤

m m+1

N mesh(C),

where m is the combinatorial dimension of C. Consequently, the mesh of C (N ) converges to 0 as N goes to infinity. Proposition 3.4.7 Let C be a simplicial complex of Rn . Then one has dim(|C|) ≤ dimcomb (C). Proof Denote by V (N ) the vertex set of the N -th barycentric subdivision C (N ) of C. By Proposition 3.3.2, the family α N := (StC (N ) ( p)) p∈V (N ) is a finite open cover

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3 Topological Dimension of Polyhedra

of |C (N ) | = |C| whose order is m := dimcomb (C). On the other hand, we have that mesh(α N ) ≤ 2 mesh(C (N ) ) by Proposition 3.4.3. By using Proposition 3.4.6, we deduce that lim N →∞ mesh(α N ) = 0. This implies dim(|C|) ≤ m by Proposition 1.4.4. Corollary 3.4.8 Every polyhedron P satisfies dim(P) < ∞.

3.5 The Lebesgue Lemma and its Applications Let n be a non-negative integer. Lemma 3.5.1 Every finite union of affine hyperplanes of Rn has empty interior in Rn . Proof This is an immediate consequence of the Baire theorem since every affine hyperplane of Rn has empty interior. Consider the unit cube [0, 1]n ⊂ Rn . For each k ∈ {1, . . . , n}, the kth lower face Fk (0) and the kth upper face Fk (1) of [0, 1]n are defined by Fk (0) := {(x1 , . . . , xn ) ∈ [0, 1]n | xk = 0}, Fk (1) := {(x1 , . . . , xn ) ∈ [0, 1]n | xk = 1}

(3.5.1) (3.5.2)

respectively. One says that the faces Fk (0) and Fk (1) are opposite faces of the cube [0, 1]n . Lemma 3.5.2 (Lebesgue’s lemma) Let α be a finite open cover of the unit cube [0, 1]n . Suppose that there is no element of α meeting two opposite faces of [0, 1]n . Then one has D(α) ≥ n. Proof We have to show that every finite open cover of [0, 1]n that is finer than α has order at least n. Suppose for contradiction that β = (Ui )i∈I is a finite open cover of [0, 1]n finer than α such that ord(β) ≤ n − 1. For each i ∈ I , let p(i) = ( pk (i))1≤k≤n ∈ {0, 1}n be the vertex of the cube [0, 1]n defined by pk (i) :=

1 if Ui meets the face Fk (0), 0 otherwise.

(3.5.3)

Let ( f i )i∈I be a family of continuous maps f i : [0, 1]n → [0, 1] satisfying the following conditions:

3.5 The Lebesgue Lemma and its Applications

61

n (C1) i∈I f i (x) = 1 for all x ∈ [0, 1] , / Ui (C2) f i (x) = 0 if x ∈ (cf. Sect. 4.2). To get such a family, we can for example choose a metric on [0, 1]n that is compatible with the topology and take f i (x) :=

gi (x) , j∈I g j (x)

where gi (x) = dist(x, [0, 1]n \Ui ). Consider now the map ϕ : [0, 1]n → [0, 1]n defined by ϕ(x) :=

f i (x) p(i).

i∈I

The point ϕ(x) belongs to the simplex whose vertex set is { p(i) | x ∈ Ui }. As ord(β) ≤ n − 1 by our hypothesis, it follows that the image set of ϕ is contained in a finite union of simplices, each of combinatorial dimension at most n − 1. By applying Lemma 3.5.1, we deduce that the image set of ϕ has empty interior in Rn . In particular, we can find a point ω in the interior of the cube [0, 1]n that does not belong to the image of ϕ. On the other hand, the fact that β is finer than α implies that no element of β meets two opposite faces of [0, 1]n . Now observe that it follows from Formula (3.5.3) that if the open set Ui meets a face of [0, 1]n , then the vertex p(i) belongs to the opposite face. Therefore, the image under ϕ of a arbitrary face of [0, 1]n is always contained in the opposite face. Consider now the radial projection π : [0, 1]n \{ω} → [0, 1]n that sends each point x ∈ [0, 1]n \{ω} to the intersection point of the half-line starting from ω and passing through x with the boundary ∂[0, 1]n = [0, 1]n \(0, 1)n of the cube [0, 1]n (see Fig. 3.3). Then the composite map ψ := π ◦ ϕ : [0, 1]n → [0, 1]n is continuous since π and ϕ are continuous. The points belonging to the interior of [0, 1]n are sent by ψ in the boundary of [0, 1]n . On the other hand, as ψ and ϕ coincide on the boundary of [0, 1]n , the image under ψ of each face of [0, 1]n is contained in the opposite face. It follows that there is no point of [0, 1]n that is fixed by ψ. This gives a contradiction since, by the Brouwer fixed point theorem, every continuous map from [0, 1]n into itself has at least one fixed point.

x ω

Fig. 3.3 The radial projection π

(x)

62

3 Topological Dimension of Polyhedra

Let 1 ≤ p ≤ ∞. Recall that the p-norm · p on Rn is defined by x p =

n

1/ p |xk |

p

k=1

if 1 ≤ p < ∞, and x ∞ = max |xk |. 1≤k≤n

Corollary 3.5.3 Denote by d p the metric on [0, 1]n associated with the norm · p (1 ≤ p ≤ ∞). Let α be a finite open cover of [0, 1]n . Suppose that the mesh of α, with respect to the metric d p , is less than 1. Then one has D(α) ≥ n. Proof No element of the cover α meets two opposite faces of [0, 1]n since the d p distance between two opposite faces is 1. Theorem 3.5.4 One has dim([0, 1]n ) = n for all n ∈ N. Proof Let ⊂ Rn be an n-simplex. We have that dim() ≤ n by Proposition 3.4.7. As [0, 1]n is homeomorphic to , we deduce that dim([0, 1]n ) ≤ n. Consider now a finite open cover α of [0, 1]n such that no element of α meets two opposite faces of [0, 1]n . We can take for example the cover α consisting of the 2n open subsets of the form U1 × · · · × Un , where Uk = [0, 1) or Uk = (0, 1] for all k = 1, . . . , n. Then α satisfies D(α) ≥ n by Lemma 3.5.2. This shows that dim([0, 1]n ) ≥ n. Corollary 3.5.5 Let C be a simplicial complex of Rn . Then one has dim(|C|) = dimcomb (C). Proof The inequality dim(|C|) ≤ dimcomb (C) follows from Proposition 3.4.7. Let m := dimcomb (C) and let ∈ C be an m-simplex. As is homeomorphic to the cube [0, 1]m , we have that dim() = m by Theorem 3.5.4. This implies dim(|C|) ≥ m by Proposition 1.2.1. Remark 3.5.6 We can also obtain Corollary 3.5.5 by applying Corollary 1.2.6 which gives dim(|C|) = max dim() = m, ∈C

since every k-simplex satisfies dim() = k by Theorem 3.5.4. Corollary 3.5.7 One has dim(Rn ) = n for all n ∈ N.

3.5 The Lebesgue Lemma and its Applications

63

Proof We have that Rn =

∞

Fk ,

k=1

where Fk := [−k, k]n . As dim(Fk ) = dim([0, 1]n ) = n, we deduce that dim(Rn ) = n by applying Theorem 1.7.1. Corollary 3.5.8 If A is an affine subspace of Rn (n ∈ N) then one has dim(A) = dima f f (A). Proof This is an immediate consequence of Corollary 3.5.7. Indeed, if A is an affine subspace of Rn then A is clearly homeomorphic to Rm , where m := dima f f (A). Corollary 3.5.9 The Hilbert cube [0, 1]N satisfies dim([0, 1]N ) = ∞. Proof The space [0, 1]n can be embedded as a closed subset of [0, 1]N so that n = dim([0, 1]n ) ≤ dim([0, 1]N ) for every n ∈ N by Proposition 1.2.1.

Corollary 3.5.10 Let P and Q be two polyhedra that are not both empty. Then one has dim(P × Q) = dim(P) + dim(Q). Proof As P is a polyhedron, we can find an integer p ≥ 0 and a simplicial complex C of R p such that P is homeomorphic to |C|. Similarly, we can find an integer q ≥ 0 and a simplicial complex C of Rq such that Q is homeomorphic to |C |. Then we have and |C | = . |C| = ∈C

∈C

Consequently, the product space P × Q is homeomorphic to |C| × |C | =

× ⊂ R p × Rq = R p+q .

∈C, ∈C

Let ∈ C be a k-simplex and ∈ C a k -simplex (k, k ∈ N). Then × is homeomorphic to [0, 1]k × [0, 1]k = [0, 1]k+k . It follows that dim( × ) = k + k by Theorem 3.5.4. On the other hand, × is compact and hence closed in |C| × |C |. By applying Corollary 1.2.6, we obtain dim(|C| × |C |) = m + m , where m (resp. m ) is the combinatorial dimension of C (resp. C ). By using

64

3 Topological Dimension of Polyhedra

Corollary 3.5.5, we finally get dim(|C| × |C |) = dim(|C|) + dim(|C |). Thus, we have that dim(P × Q) = dim(P) + dim(Q). The above proof obviously extends to a finite product of polyhedra. Thus, we have also the following: Corollary 3.5.11 Let P1 , P2 , . . . , Pn be a finite sequence of non-empty polyhedra. Then one has dim(P1 × P2 × · · · × Pn ) = dim(P1 ) + dim(P2 ) + · · · + dim(Pn ).

3.6 Abstract Simplicial Complexes Definition 3.6.1 An abstract simplicial complex is a pair = (V, ) consisting of a finite set V and of a set whose elements are subsets of V satisfying the following condition: (ASC) if σ ∈ and σ ⊂ σ, then one has σ ∈ . Let = (V, ) be an abstract simplicial complex. The elements of V are called the vertices of and the elements of are called the simplices of . The combinatorial dimension dimcomb (σ) ∈ {−1} ∪ N of a simplex σ ∈ is defined by dimcomb (σ) := −1 + #σ. The combinatorial dimension dimcomb () ∈ {−1} ∪ N of is defined by dimcomb () := max dimcomb (σ). σ∈

Example 3.6.2 Take V := {1, 2, 3, 4} and := {∅, {1}, {2}, {3}, {4}, {1, 2}, {2, 3}, {1, 3}, {2, 4}, {1, 2, 3}}. One immediately checks that (ASC) is satisfied. Therefore := (V, ) is an abstract simplicial complex. It has combinatorial dimension dimcomb () = 2. Example 3.6.3 Let C be a simplicial complex of Rn . Let V be the set of vertices of C and let be the set consisting of all σ ⊂ V such that there exists a simplex ∈ C whose set of vertices is σ. Condition (SC1) in Definition 3.2.1 implies that := (V, ) satisfies (ASC). One says that is the abstract simplicial complex associated with C. Note that the combinatorial dimension of is equal to the combinatorial dimension of C and hence, by Corollary 3.5.5, to the topological dimension dim(|C|) of its support |C| ⊂ Rn .

3.6 Abstract Simplicial Complexes

65

Two abstract simplicial complexes = (V, ) and = (V , ) are called isomorphic if there exists a bijective map f : V → V such that σ ∈ ⇐⇒ f (σ) ∈ for all σ ⊂ V . Let = (V, ) be an abstract simplicial complex. Let (ev )v∈V be the canonical basis of RV (we may identify RV with Rn , for n = #V ). For each σ ∈ , denote by σ the simplex in RV whose vertex set is {ev | v ∈ σ}. Clearly, we have that σ ∩ σ = σ∩σ for all σ, σ ∈ . It follows that C := {σ | σ ∈ } is a simplicial complex of RV . One says that C is the geometric realization of the abstract simplicial complex . Note that the combinatorial dimension of is equal to the combinatorial dimension of C. Observe also that the abstract simplicial complex associated with C is the abstract simplicial complex 0 = (V0 , ), where V0 ⊂ V is the set of active vertices of , that is, the set of v ∈ V such that {v} ∈ .

Notes The Lebesgue lemma (cf. Lemma 3.5.2) about the finite open covers of the n-cube was stated by Lebesgue in [66]. It was used by Lebesgue in order to show that [0, 1]n and [0, 1]m are not homeomorphic for n = m. The proof of Lebesgue’s lemma given in [66] contains an error as was pointed out by Brouwer in [19]. A corrected proof appeared in [67]. Let D n (resp. Sn−1 ) denote the closed unit ball (resp. the unit sphere) in the Euclidean space Rn . The Brouwer fixed point theorem, which says that every continuous map f : D n → D n admits at least one fixed point, may be proved by using tools from algebraic topology (see for example [101]) in the following way. Suppose for contradiction that there exists a continuous map f : D n → D n without fixed points (n ≥ 2). Consider the map g : D n → S n−1 that sends every point x ∈ D n to the intersection point of the half-line starting from f (x) and passing through x with the boundary sphere Sn−1 (see Fig. 3.4). Clearly g is continuous. On the other hand, the map g fixes every point belonging to Sn−1 , that is, it satisfies g ◦ h = i,

(3.6.1)

where h is the inclusion map Sn−1 → D n and i is the identity map on Sn−1 . On the level of n − 1-dimensional real homology, the maps g, h and i induce linear

66

3 Topological Dimension of Polyhedra

f(x)

x g(x)

Fig. 3.4 The map g : D n → S n−1

maps g∗ : Hn−1 (D n ) → Hn−1 (Sn−1 ), h ∗ : Hn−1 (Sn−1 ) → Hn−1 (Sn−1 ) and i ∗ = Id : Hn−1 (Sn−1 ) → Hn−1 (Sn−1 ), where Id is the identity map on Hn−1 (Sn−1 ). Now it follows from (3.6.1) that g∗ ◦ h ∗ = I∗ = Id, which is impossible since Hn−1 (D n ) = 0 while Hn−1 (Sn−1 ) ∼ = R = 0. Actually, there are many other proofs of the Brouwer fixed point theorem (see [116] and the references therein). The one presented by Milnor in [76] is elementary an especially clever.

Exercises 3.1 Show that, up to homeomorphism, there are only countably many polyhedra. 3.2 Let X be a non-empty topological space. Show that X is a polyhedron with dim(X ) = 0 if and only if X is finite and discrete. 3.3 Show that a polyhedron has only finitely many connected components. 3.4 A topological space X is called path-connected if, given any two points x, y ∈ X , there exists a continuous map γ : [0, 1] → X such that γ(0) = x and γ(1) = y. Show that every connected polyhedron is path-connected. 3.5 Show that neither the Cantor set nor the Hilbert cube are polyhedra. 3.6 Find a proof of Lemma 3.5.1 that does not require Baire’s theorem. Hint: proceed by contradiction and use induction on the dimension n (consider a suitable translate of one of the hyperplanes for going from n to n − 1). 3.7 (Lebesgue’s lemma for closed coverings). Let α be a finite closed cover of [0, 1]n such that no element of α meets two opposite faces of [0, 1]n . Show that one has ord(α) ≥ n. Hint: use Proposition 1.6.3 and Lemma 3.5.2. 3.8 Show that the unit sphere Sn−1 ⊂ Rn , defined by Sn−1 := {(x1 , . . . , xn ) ∈ Rn |

n k=1

satisfies dim(Sn−1 ) = n − 1.

xk2 = 1},

Exercises

67

3.9 Let X be a polyhedron and n a non-negative integer. Show that one has dim(X ) ≥ n if and only if X contains a subset homeomorphic to [0, 1]n . 3.10 Let = (V, ) and = (V , ) be two abstract simplicial complexes. Equip each of the sets V and V with some total ordering. Let W := V × V denote the Cartesian product of the sets V and V . Consider the set consisting of all subsets λ ⊂ W satisfying the following condition: there exist an integer n ≥ 0, vertices v0 , v1 , . . . , vn ∈ V and v0 , v1 , . . . , vn ∈ V such that v0 ≤ v1 ≤ · · · ≤ vn , v0 ≤ v1 ≤ · · · ≤ vn , {v0 , v1 , . . . , vn } ∈ , {v0 , v1 , . . . , vn } ∈ , and λ = {(v0 , v0 ), (v1 , v1 ), . . . , (vn , vn )}. (a) Show that := (W, ) is an abstract simplicial complex. (b) Let P (resp. P , resp. Q) denote the support of the geometric realization of (resp. , resp. ). Show that Q is homeomorphic to the topological product P × P . 3.11 Let X and Y be polyhedra. Show that the product space X × Y is a polyhedron. Hint: use Exercise 3.10.

Chapter 4

Dimension and Maps

In this chapter, we establish Urysohn’s lemma (Lemma 4.1.2) and the Tietze extension theorem (Theorem 4.1.4) for normal spaces. We introduce the notion of ε-injective map and prove the theorem of Alexandroff saying that a compact metric space X satisfies dim(X ) ≤ n if and only if for every ε > 0, there exists a polyhedron P such that dim(P) ≤ n and an ε-injective continuous map from X into P (Theorem 4.5.4). We deduce that dim(X × Y ) ≤ dim(X ) + dim(Y ) whenever X and Y are compact metrizable spaces that are not both empty (Corollary 4.5.6). We also prove the Nöbeling-Pontryagin embedding theorem (Corollary 4.7.6) asserting that every compact metrizable space with topological dimension n embeds in R2n+1 .

4.1 The Tietze Extension Theorem Lemma 4.1.1 Let X be a topological space. Let A and B be disjoint closed subsets of X . Suppose that there exists a dense subset E of the unit segment [0, 1] and a family ((t))t∈E of open subsets of X satisfying A ⊂ (t) ⊂ (t) ⊂ (t ) ⊂ X \B

(4.1.1)

for all t, t ∈ E such that t < t . Then there exists a continuous map f : X → [0, 1] such that f (a) = 0 for all a ∈ A and f (b) = 1 for all b ∈ B. Proof For each x ∈ X , let E(x) denote the set consisting of all t ∈ E such that x ∈ (t). Consider the map f : X → [0, 1] defined by f (x) :=

inf E(x) 1

if E(x) = ∅, if E(x) = ∅.

We deduce from (4.1.1) that f (x) = 0 for all x ∈ A and f (x) = 1 for all x ∈ B. © Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_4

69

70

4 Dimension and Maps

Let us fix now x0 ∈ X and ε > 0. If t1 ∈ E satisfies f (x0 ) < t1 < f (x0 ) + ε, then (t1 ) is an open neighborhood of x0 , and we have that f (x) < f (x0 ) + ε for all x ∈ (t1 ). On the other hand, if t2 ∈ E satisfies f (x0 ) − ε < t2 < f (x0 ), then X \(t2 ) is an open neighborhood of x0 , and we have that f (x0 ) − ε < f (x) for all x ∈ X \(t2 ). We conclude that f is continuous at x0 . Lemma 4.1.2 (Urysohn lemma) Let X be a normal space. Let A and B be disjoint closed subsets of X . Then there exists a continuous map f : X → [0, 1] such that f (a) = 0 for all a ∈ A and f (b) = 1 for all b ∈ B. Proof Let

E :=

k | n ∈ N, k ∈ {0, 1, . . . , 2n } 2n

denote the set consisting of all dyadic rationals in [0, 1]. We construct a family ((t))t∈E of open subsets of X satisfying the conditions of Lemma 4.1.1 in the following way. We first take (1) := X \B. As X is normal, it follows from Proposition 1.5.2 that we can find an open subset (0) ⊂ X such that A ⊂ (0) ⊂ (0) ⊂ (1). Suppose now that we have already constructed, for some integer n ≥ 0 and all k k ∈ {0, 1, . . . , 2n }, open subsets n ⊂ X such that 2

k 2n

⊂

k+1 2n

for all k ∈ {0, 1, . . . , 2n − 1}. By applying again Proposition 1.5.2, we can find, for 2k + 1 n ⊂ X such that every k ∈ {0, 1, . . . , 2 − 1}, an open subset 2n+1

k n 2

2k + 1 ⊂ 2n+1

2k + 1 ⊂ 2n+1

k+1 ⊂ . 2n

By induction on n, we construct in this way a family ((t))t∈E of open subsets of X with the required properties. As the set E of dyadic rationals is dense in [0, 1], it follows from Lemma 4.1.1 that there exists a continuous map f : X → [0, 1] such that f (a) = 0 for all a ∈ A and f (b) = 1 for all b ∈ B. Lemma 4.1.3 Let X be a normal space and let Y be a closed subset of X . Let f : Y → R be a continuous map. Suppose that there is a constant C ≥ 0 such that | f (y)| ≤ C for all y ∈ Y . Then there exists a continuous map g : X → R such that |g(x)| ≤ C/3 for all x ∈ X and | f (y) − g(y)| ≤ 2C/3 for all y ∈ Y .

4.1 The Tietze Extension Theorem

71

Proof Consider the sets A := {y ∈ Y | f (y) ≤ −C/3} and B := {y ∈ Y | f (y) ≥ C/3} Clearly A and B are disjoint closed subsets of X . Therefore, it follows from Lemma 4.1.2 that there exists a continuous map h : X → [0, 1] such that h(a) = 0 for all a ∈ A and h(b) = 1 for all b ∈ B. Then the map g : X → R defined by g(x) := C(2h(x) − 1)/3 for all x ∈ X is also continuous. Moreover, it satisfies |g(x)| ≤ C/3 for all x ∈ X since 0 ≤ h(x) ≤ 1. To complete the proof, it remains only to show that | f (y) − g(y)| ≤ 2C/3 (4.1.2) for all y ∈ Y . Suppose first that y ∈ A. Then we have that −C ≤ f (y) ≤ −C/3 and g(y) = −C/3 so that (4.1.2) is satisfied. Similarly, for y ∈ B, we have that C/3 ≤ f (y) ≤ C and g(y) = C/3 so that (4.1.2) also holds. Finally, if y ∈ Y \(A ∪ B), then | f (y)| < C/3 and |g(y)| ≤ C/3 so that (4.1.2) remains true by the triangle inequality. Let X and Z be sets and let Y ⊂ X . Let f : Y → Z be a map. One says that a map F : X → Z extends f , or that F is an extension of f , if one has F(y) = f (y) for all y ∈ Y . Theorem 4.1.4 (Tietze extension theorem) Let X be a normal space and let Y be a closed subset of X . Let I ⊂ R be any interval and suppose that f : Y → I is a continuous map. Then there exists a continuous map F : X → I extending f . Proof We may assume that I contains more than one point since otherwise the statement is trivial. We shall distinguish three cases depending on the homeomorphism type of the interval I . Suppose first that I is a segment, i.e., I = [α, β] with α < β. As all segments in R are homeomorphic, we can assume, without loss of generality, that I = [−C, C] for some C > 0. By Lemma 4.1.3, we can find a continuous map g0 : X → R such that |g0 (x)| ≤ C/3 for all x ∈ X and | f (y) − g0 (y)| ≤ 2C/3 for all y ∈ Y . By applying again Lemma 4.1.3, with f replaced by f − g0 |Y and C replaced by 2C/3, we can find a continuous map g1 : X → R such that |g1 (x)| ≤ 2C/9 for all x ∈ X and | f (y) − g0 (y) − g1 (y)| ≤ 4C/9 for all y ∈ Y . Continuing in this way, we obtain by induction a sequence of continuous maps gn : X → R satisfying, for every n ∈ N, the following conditions: (C1) |gn (x)| ≤ 2n C/3n+1 for all x ∈ X ; (C2) | f (y) − g0 (y) − g1 (y) − · · · − gn (y)| ≤ 2n+1 C/3n+1 for all y ∈ Y . in f t y n n+1 = C < ∞, we deduce from (C1) that the series As ∞ n=0 2 C/3 n=0 gn (x) is normally convergent and hence uniformly convergent on X . Moreover, we have that n+1 C/3n+1 tends to 0 as | ∞ n=0 gn (x)| ≤ C for all x ∈ X . On the other hand, since 2 n goes to infinity, it follows from (C2) that the continuous map F : X → [−C, C],

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defined by F(x) := ∞ n=0 gn (x) for all x ∈ X , satisfies F(y) = f (y) for all y ∈ Y . Thus F has the required properties. Suppose now that I is an half-open interval. We can assume I = [0, 1). By the first part of the proof, there exists a continuous map G : X → [0, 1] such that G(y) = f (y) for all y ∈ Y . Consider the subset A ⊂ X defined by A := G −1 (1). As A and Y are disjoint closed subsets of X , it follows from Lemma 4.1.2 that there exists a continuous map ϕ : X → [0, 1] such that ϕ(a) = 0 for all a ∈ A and ϕ(y) = 1 for all y ∈ Y . Then the map F : X → [0, 1) defined by F(x) := ϕ(x)G(x) is continuous and extends f . Finally, suppose that I is an open interval (e.g., I = R). In that case, we can assume I = (0, 1). By the second part of the proof, there exists a continuous map H : X → (0, 1] such that h(y) = f (y) for all y ∈ Y . We then use the same kind of argument, namely we apply Lemma 4.1.2 again to get a continuous map ψ : X → [0, 1] such that ψ(a) = 0 for all a ∈ A := H −1 (1) and ψ(y) = y for all y ∈ Y . Then the map F : X → (0, 1), defined by F(x) := ψ(x)H (x) for all x ∈ X , has the required properties. Let X be a compact space and let C(X ) denote the vector space consisting of all continuous maps f : X → R. Recall that C(X ) is a Banach space for the sup-norm, X defined by i.e., the norm · ∞ X := sup | f (x)|. f ∞ x∈X

Corollary 4.1.5 Let X be a compact Hausdorff space and let Y be a closed subset of X . Suppose that f : Y → R is a continuous map. Then there exists a continuous X = f Y . map F : X → R extending f such that F∞ ∞ Proof As every compact Hausdorff space is normal by Proposition 1.5.4, this immeY , f Y ]. diately follows from Theorem 4.1.4 by taking I = [− f ∞ ∞

4.2 Partitions of Unity Let X be a topological space. The support of a map f : X → R is the closure of the set of points in X where f does not vanish, i.e., the closed subset supp( f ) ⊂ X defined by supp( f ) := {x ∈ X | f (x) = 0}. Note that the set X \ supp( f ) is the largest open subset of X on which the map f is identically zero. Let α = (Ui )i∈I be a finite open cover of X . One says that a family ( f i )i∈I of continuous maps f i : X → [0, 1] is a partition of unity subordinate to α if it satisfies the following conditions: (Pu-1) i∈I f i (x) = 1 for all x ∈ X ; (Pu-2) supp( f i ) ⊂ Ui for all i ∈ I .

4.2 Partitions of Unity

73

Proposition 4.2.1 Let X be a normal space and let α = (Ui )i∈I be a finite open cover of X . Then there exists a partition of unity subordinate to α. Proof By Corollary 1.6.4, there exists an open cover (Vi )i∈I of X such that Vi ⊂ Ui for all i ∈ I . By applying again Corollary 1.6.4, we can find an open cover (Wi )i∈I of X satisfying Wi ⊂ Vi for all i ∈ I . Now, by Lemma 4.1.2, there exists, for each i ∈ I , a continuous map gi : X → [0, 1] that takes the value 1 at every point of Wi and the value 0 at every point of X \Vi . The support of gi is contained in Vi and hence in Ui . On the other hand, we have that

gi (x) > 0

i∈I

for every x ∈ X since the sets Wi , i ∈ I , cover X . It follows that the family ( f i )i∈I , where f i : X → [0, 1] is defined by f i (x) :=

gi (x) j∈I g j (x)

for all x ∈ X , is a partition of unity subordinate to α.

4.3 Nerve of a Cover Let α = (Ui )i∈I be a finite family of subsets of a set X . The nerve of α is the abstract simplicial complex N whose set of vertices is I and whose simplices are the subsets J ⊂ I such that Ui = ∅. i∈J

It is obvious that N satisfies condition (ASC) of Definition 3.6.1. Observe that the simplicial dimension of the nerve N is equal to the order of α. Suppose now that α is a finite open cover of a topological space X and that ( f i )i∈I is a partition of unity subordinate to α. Let C denote the simplicial complex of R I that is the geometric realization of the nerve N of α. Consider, for each x ∈ X , the point f (x) ∈ R I defined by f (x) :=

f i (x)ei ,

i∈I I where (ei )i∈I is the canonical basis

of R . Denote by I (x) the set consisting of all i ∈ I such that x ∈ Ui . As x ∈ i∈I (x) Ui , the subset I (x) ⊂ I is a simplex of N . We have that f i (x)ei and f i (x) = 1. f (x) = i∈I (x)

i∈I (x)

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This shows that f (x) belongs to the simplex I (x) ∈ C. It follows that f (x) ∈ |C| for all x ∈ X . One says that the continuous map f : X → |C| is the map associated with the finite open cover α and the partition of unity ( f i )i∈I . We shall use the following observation in the next section. Proposition 4.3.1 With the above notation, for every vertex p = ei of C, one has f −1 StC ( p) ⊂ Ui . (Recall that StC ( p) denote the open star of C at p as defined in Sect. 3.3.) Proof Let x ∈ X \Ui . The point f (x) belongs to the simplex of C whose vertices are the points e j such that x ∈ U j . As the point p = ei is not one of the vertices of this simplex, we have that f (x) ∈ |C|\ StC ( p). This shows that the inverse image of StC ( p) by f is contained in Ui .

4.4 α-Compatible Maps Let f : X → Y be a map from a set X into a set Y . The inverse image of a family β = (Bi )i∈I of subsets of Y is the family f −1 (β) of subsets of X defined by f −1 (β) := ( f −1 (Bi ))i∈I . Proposition 4.4.1 Let f : X → Y be a map from a set X to a set Y . Let β = (Bi )i∈I be a family of subsets of Y . Then one has (i) (ii) (iii) (iv)

ord x ( f −1 (β)) = ord f (x) (β) for every x ∈ X ; ord( f −1 (β)) ≤ ord(β); ord( f −1 (β)) = ord(β) if f is surjective; f −1 (β) is a cover of X if β is a cover of Y .

Proof By definition, we have that x ∈ f −1 (Bi ) if and only if f (x) ∈ Bi . It follows that ord x ( f −1 (β)) = −1 + #{i ∈ I | x ∈ f −1 (Bi )} = −1 + #{i ∈ I | f (x) ∈ Bi } = ord f (x) (β).

This shows (i). Properties (ii), (iii), and (iv) are immediate consequences of (i). Suppose now that X and Y are topological spaces and that f : X → Y is a continuous map. If β is an open cover (resp. a closed cover) of Y , then f −1 (β) is an open cover (resp. a closed cover) of X . Proposition 4.4.2 Let X and Y be topological spaces and let f : X → Y be a continuous map. Let β be a finite open cover of Y . Then one has D( f −1 (β)) ≤ D(β).

4.4 α-Compatible Maps

75

Proof Let γ be a finite open cover of Y such that γ β and ord(γ) = D(β). Then we have that f −1 (γ) f −1 (β) and hence D( f −1 (β)) ≤ ord( f −1 (γ)). As ord( f −1 (γ)) ≤ ord(γ) by Proposition 4.4.1(ii), we deduce that D( f −1 (β)) ≤ ord(γ) = D(β).

Definition 4.4.3 Let X and Y be topological spaces. Let α be a finite open cover of X . A continuous map f : X → Y is said to be α-compatible if there exists a finite open cover β of Y such that f −1 (β) α. Proposition 4.4.4 Let α = (Ui )i∈I be a finite open cover of a topological space X . Let C denote the geometric realization of the nerve of α. Suppose that ( f i )i∈I is a partition of unity subordinate to α. Then the map f : X → |C| associated with the cover α and the partition of unity ( f i )i∈I is α-compatible. Proof Let V denote the set of vertices of C. Then β := (StC (v))v∈V is a finite open cover of |C| by Proposition 3.3.2. As f −1 (β) α by Proposition 4.3.1, it follows that f is α-compatible. Proposition 4.4.5 Let X be a topological space and α a finite open cover of X . Suppose that there exist a topological space Y and an α-compatible continuous map f : X → Y . Then one has D(α) ≤ dim(Y ). Proof As f is α-compatible, there exists a finite open cover β of Y such that f −1 (β) α. We have that D(α) ≤ D( f −1 (β)) by Proposition 1.1.4. Since D( f −1 (β)) ≤ D(β) by Proposition 4.4.2, we deduce that D(α) ≤ D(β) ≤ dim(Y ). Proposition 4.4.6 Let X be a normal space. Let α be a finite open cover of X . Then there exists a polyhedron P with topological dimension dim(P) = D(α) and an α-compatible continuous map f : X → P. Proof Let β be a finite open cover of X such that β α and ord(β) = D(α). Let C denote the geometric realization of the nerve N of β. Consider the polyhedron P := |C|. By Proposition 4.2.1, we can find a partition of unity ( f i )i∈I subordinate to β. Let f : X → P denote the map associated with the cover β and the partition of unity ( f i )i∈I . By Corollary 3.5.5, the topological dimension dim(P) of P is equal to the simplicial dimension of N and hence to the order of β. Consequently, we have that dim(P) = ord(β) = D(α). On the other hand, the map f is βcompatible by Proposition 4.4.4. As the cover β is finer than α, it follows that f is also α-compatible. Theorem 4.4.7 Let X be a normal space. Let n ∈ N. Then the following conditions are equivalent: (a) dim(X ) ≤ n;

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(b) for every finite open cover α of X , there exist a polyhedron P such that dim(P) ≤ n and an α-compatible continuous map f : X → P; (c) for every finite open cover α of X , there exist a topological space Y such that dim(Y ) ≤ n and an α-compatible continuous map f : X → Y . Proof The fact that (a) implies (b) immediately follows from Proposition 4.4.6 since every finite open cover α of X satisfies D(α) ≤ dim(X ). Condition (b) trivially implies (c). Finally, Proposition 4.4.5 shows us that (a) is a consequence of (c) since, by definition, dim(X ) = sup D(α), where α runs over all finite open covers of X .

4.5 ε-Injective Maps Let (X, d) be a metric space and Y a set. Given a real number ε > 0, a map f : X → Y is called ε-injective if it satisfies f (x1 ) = f (x2 ) ⇒ d X (x1 , x2 ) < ε for all x1 , x2 ∈ X . Remark 4.5.1 The map f is injective if and only if it is ε-injective for every ε > 0. Remark 4.5.2 If diam( f −1 (y)) < ε for all y ∈ Y , then the map f is ε-injective. Observe that the converse is also true if we assume that X is compact and that Y is equipped with a Hausdorff topology such that f is continuous. Lemma 4.5.3 Let X be a compact space and Y a Hausdorff space. Let α = (Ui )i∈I be an open cover of X . Let f : X → Y be a continuous map satisfying the following condition: for every y ∈ Y , there exists i ∈ I such that f −1 (y) ⊂ Ui (in other words, the closed cover γ of X defined by γ = ( f −1 (y)) y∈Y is finer than α). Then there exists an open cover β = (Vi )i∈I of Y such that f −1 (Vi ) ⊂ Ui for all i ∈ I . Proof Consider, for each i ∈ I , the subset Vi ⊂ Y defined by Vi := {y ∈ Y | f −1 (y) ⊂ Ui }. Then we clearly have that f −1 (Vi ) ⊂ Ui . On the other hand, it follows from our hypothesis on f that the sets Vi cover Y . Finally, observe that Vi = Y \Fi , where Fi = f (X \Ui ). As X \Ui is compact and f is continuous, the set Fi is compact. Since Y is Hausdorff, it follows that Fi is a closed subset of Y . Thus Vi is open in Y for every i ∈ I . This shows that the family β := (Vi )i∈I has the required properties. The following theorem gives a characterization of the topological dimension of a compact metric space in terms of ε-injective maps.

4.5 ε-Injective Maps

77

Theorem 4.5.4 Let X be a compact metric space. Let n ∈ N. Then the following conditions are equivalent: (a) dim(X ) ≤ n; (b) for every ε > 0, there exists a polyhedron P with dim(P) ≤ n and an ε-injective continuous map f : X → P; (c) for every ε > 0, there exists a compact metrizable space Y with dim(Y ) ≤ n and an ε-injective continuous map f : X → Y ; (d) for every ε > 0, there exists a Hausdorff space Y with dim(Y ) ≤ n and an ε-injective continuous map f : X → Y . Proof Suppose that dim(X ) ≤ n. Let ε > 0. Consider the cover of X formed by its open balls of radius ε/2. By compactness of X , this cover admits a finite subcover α. By Theorem 4.4.7, we can find a polyhedron P with dim(P) ≤ n and an αcompatible continuous map f : X → P. Then f is ε-injective. This shows that (a) implies (b). Condition (b) clearly implies (c) since any polyhedron is compact and metrizable. Also (c) implies (d) since any metrizable space is Hausdorff. Finally, let us show that (d) implies (a). Suppose (d). Consider a finite open cover α = (Ui )i∈I of X . Let λ > 0 be a Lebesgue number for α. By (d), we can find a Hausdorff space Y with dim(Y ) ≤ n and a λ-injective continuous map f : X → Y . For every y ∈ Y , we have that diam( f −1 (y)) ≤ λ. Therefore, there exists i ∈ I such that f −1 (y) ⊂ Ui . Thus, it follows from Lemma 4.5.3 that f is α-compatible. We then deduce that D(α) ≤ n by applying Proposition 4.4.5. Consequently, X satisfies (a). Remark 4.5.5 We cannot remove the hypothesis saying that Y is Hausdorff in condition (d) of Theorem 4.5.4. Indeed, for every topological space X , there exists a topological space Y with dim(Y ) = 0 and an injective continuous map f : X → Y (we can take for example as Y the set underlying X equipped with the trivial topology and as f the identity map). Corollary 4.5.6 Let X and Y be compact metrizable spaces that are not both empty. Then one has dim(X × Y ) ≤ dim(X ) + dim(Y ). (4.5.1) Proof We may assume 0 ≤ dim(X ) < ∞ and 0 ≤ dim(Y ) < ∞. Let d X and dY be metrics on X and Y respectively that are compatible with the topologies. Let us equip X × Y with the metric defined by d((x, y), (x , y )) := max(d X (x, x ), dY (y, y )) for all (x, y), (x , y ) ∈ X ×Y . The metric d is compatible with the product topology on X × Y . Let ε > 0. By Theorem 4.5.4, we can find a polyhedron P with dim(P) ≤ dim(X ) and an ε-injective continuous map f : X → P. Similarly, we can find a polyhedron Q with dim(Q) ≤ dim(Y ) and an ε-injective continuous map g : Y → Q.

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Then the product map F : X × Y → P × Q, defined by F(x, y) := ( f (x), g(y)) for all (x, y) ∈ X × Y , is clearly ε-injective. As P and Q are polyhedra, we have that dim(P × Q) = dim(P) + dim(Q) by Corollary 3.5.10. We deduce that dim(P × Q) ≤ dim(X ) + dim(Y ). As P × Q is Hausdorff, it follows from Theo rem 4.5.4 that dim(X × Y ) ≤ dim(X ) + dim(Y ). Observe that Corollary 4.5.6 implies in particular that if X and Y are compact metrizable spaces with dim(X ) = dim(Y ) = 0 then one has dim(X × Y ) = 0. This last result remains true if X and Y are only assumed to be compact and Hausdorff by Corollary 2.4.24. However, it becomes false for general topological spaces X and Y . Indeed, in Example 2.4.29, we described a non-accessible topological space X satisfying dim(X ) = 0 and dim(X × X ) = 2. In Sect. 5.5, we shall give an example of a normal Hausdorff space X such that dim(X ) = 0 and dim(X × X ) ≥ 1.

4.6 Definition of dimε (X, d) Let (X, d) be a metric space. Given a real number ε > 0, we define the quantity dimε (X, d) by dimε (X, d) := inf dim(K ), K

where K runs over all compact metrizable spaces for which there exists an ε-injective continuous map f : X → K . Example 4.6.1 If d is the usual metric on the unit segment [0, 1], then we clearly have 1 if 0 < ε ≤ 1, dimε ([0, 1], d) = 0 if ε > 1. Proposition 4.6.2 Let (X, d) be a compact metric space. Then the following hold: (a) (b) (c) (d) (e)

dimε (X, d) ≤ dim(X ) for all ε > 0; dimε (X, d) < ∞ for all ε > 0; the map ε → dimε (X, d) is non-increasing on (0, ∞); limε→0 dimε (X, d) = dim(X ); if dim(X ) < ∞, then there exists ε0 > 0 such that dimε (X, d) = dim(X ) for all 0 < ε < ε0 .

Proof Assertion (a) follows from the fact that the identity map Id X : X → X is injective and hence ε-injective for every ε > 0. Let ε > 0. Since X is compact, we can find a finite open cover α of X consisting of open balls of radius smaller than ε/2. Consider a partition of unity ( fi )i∈I subordinate to α. Let C denote the geometric realization of the nerve of α and f : X → |C| the map associated with α and the partition of unity ( f i )i∈I . As the map f is ε-injective and continuous, it follows that dimε (X, d) ≤ dim(|C|) < ∞. This shows (b).

4.6 Definition of dimε (X, d)

79

If K is a compact metrizable space and f : X → K is ε-injective, then f is also ε injective for every ε ≥ ε. Consequently, the map ε → dimε (X, d) is non-increasing. This shows (c). Let n be an integer such that n < dim(X ). By Theorem 4.5.4, there exists ε0 > 0 such that dimε0 (X, d) > n. Then for every 0 < ε < ε0 , we have that n < dimε (X, d) ≤ dim(X ). This shows (d). Assertion (e) immediately follows from (c) and (d) since dimε (X, d) is an integer for every ε > 0. The following results will be used in Sect. 7.1. Proposition 4.6.3 Let (X, d X ) and (Y, dY ) be compact metric spaces. Suppose that there exists a continuous map ϕ : X → Y such that d X (x1 , x2 ) ≤ dY (ϕ(x1 ), ϕ(x2 )) for all x1 , x2 ∈ X . Then one has dimε (X, d X ) ≤ dimε (Y, dY ) for all ε > 0. Proof It suffices to observe that if f : Y → K is ε-injective, then f ◦ ϕ : X → K is ε-injective. Corollary 4.6.4 Let d and d be metrics on a set X that define the same topology on X and such that d(x, y) ≤ d (x, y) for all x, y ∈ X . Then one has dimε (X, d) ≤ dimε (X, d ) for all ε > 0. Proof It suffices to take as ϕ the identity map on X .

Proposition 4.6.5 Let n ∈ N and p ∈ [1, ∞]. Let d be the metric on [0, 1]n ⊂ Rn induced by the norm · p . Then one has dimε ([0, 1]n , d) = n for all 0 < ε ≤ 1. Proof We have that dimε ([0, 1]n , d) ≤ dim([0, 1]n ) = n for all ε > 0 by Proposition 4.6.2(a) and Theorem 3.5.4.

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Suppose now that 0 < ε ≤ 1. Let K be a compact metrizable space and f : [0, 1]n → K a continuous map that is ε-injective with respect to the metric d. Consider the cover α of [0, 1]n whose elements are the 2n open subsets of the form U1 × · · · × Un , where Uk = [0, 1) or Uk = (0, 1] for all k ∈ {1, . . . , n}. Let y ∈ K . As ε ≤ 1, the ε-injectivity of f implies that f −1 (y) cannot meet two opposite faces of the cube [0, 1]n . Consequently, the cover ( f −1 (y)) y∈K is finer than α. By applying Lemma 4.5.3, we deduce that the map f is α-compatible. Therefore we have that D(α) ≤ dim(K ) by Proposition 4.4.5. As D(α) ≥ n by Lemma 3.5.2, we deduce that dim(K ) ≥ n. This shows that dimε ([0, 1]n , d) ≥ n.

4.7 Euclidean Embeddings of Finite-Dimensional Spaces Let m, r ∈ N. One says that a finite sequence of points p0 , p1 , . . . , pr ∈ Rm is in general position if every subsequence of cardinality ≤ m + 1 is affinely independent, i.e., for every sequence of integers 0 ≤ i 0 < i 1 < · · · < i k ≤ r , with k ≤ m, the points pi0 , pi1 , . . . , pik are affinely independent. Remark 4.7.1 If r ≤ m, the sequence p0 , p1 , . . . , pr is in general position if and only if it is affinely independent. If m ≤ r , the sequence p0 , p1 , . . . , pr is in general position if and only if every subsequence of cardinality m +1 is affinely independent. Let d be a metric on Rm compatible with the topology. Lemma 4.7.2 Let m, r ∈ N. Let p0 , p1 , . . . , pr ∈ Rm . Then, for every ε > 0, there exists a sequence of points q0 , q1 , . . . , qr ∈ Rm in general position such that d( pi , qi ) < ε for all i ∈ {0, 1, . . . , r }. In other words, the set of (r + 1)tuples (q0 , q1 , . . . , qr ) ∈ (Rm )r +1 in general position is dense in the product space (Rm )r +1 . Proof We proceed by induction on r . Suppose that there exist points q0 , q1 , . . . , qr −1 in general position in Rm such that d( pi , qi ) ≤ ε for all 0 ≤ i ≤ r − 1. Let E denote the union of all the affine subspaces of Rm that can be generated by at most m points in the set {q0 , q1 , . . . , qr −1 }. The set E is contained in a finite union of affine hyperplanes of Rm and therefore E has empty interior in Rm by Lemma 3.5.1. Consequently, we can find a point qr ∈ Rm \E such that d( pr , qr ) ≤ ε. Then the points q0 , q1 , . . . , qr are in general position. Let X be a compact Hausdorff space and Y a metric space. Let C(X, Y ) denote the space consisting of all continuous maps f : X → Y . We equip C(X, Y ) with the metric d∞ defined by d∞ ( f, g) := sup dY ( f (x), g(x)). x∈X

The topology on C(X, Y ) associated with the metric d∞ is the topology of uniform convergence.

4.7 Euclidean Embeddings of Finite-Dimensional Spaces

81

If X is a compact metric space, Y a metric space, and ε a positive real number, we denote by Cε (X, Y ) the subset of C(X, Y ) consisting of all continuous maps f : X → Y that are ε-injective. Lemma 4.7.3 Let X be a compact metric space, Y a metric space, and ε > 0. Then the set Cε (X, Y ) is open in C(X, Y ). Proof Let f ∈ Cε (X, Y ). The set K := {(x1 , x2 ) ∈ X × X | d X (x1 , x2 ) ≥ ε} is closed in X × X and hence compact. As f is ε-injective, we have that dY ( f (x1 ), f (x2 )) > 0 for all (x1 , x2 ) ∈ K . Thus, there is a real number δ > 0 such that dY ( f (x1 ), f (x2 )) ≥ δ for all (x1 , x2 ) ∈ K . Consider a continuous map g : X → Y such that d∞ ( f, g) ≤ δ/4. By applying the triangle inequality, we get, for all (x1 , x2 ) ∈ K , dY (g(x1 ), g(x2 )) ≥ dY ( f (x1 ), f (x2 )) − dY ( f (x1 ), g(x1 )) − dY ( f (x2 ), g(x2 )) ≥ δ − δ/4 − δ/4 = δ/2 > 0. It follows that g is ε-injective. This shows that Cε (X, Y ) is open in C(X, Y ).

Lemma 4.7.4 Let m, n ∈ N such that m ≥ 2n + 1. Let X be a compact metric space such that dim(X ) ≤ n and let ε > 0. Then the set Cε (X, Rm ) is dense in C(X, Rm ). Proof Let f : X → Rm be a continuous map and δ > 0. Let us show that there exists an ε-injective continuous map g : X → Rm such that d∞ ( f, g) ≤ δ. As X is compact, the map f is uniformly continuous. Thus, there exists η > 0 such that d( f (x), f (y)) ≤ δ/2 if x, y ∈ X satisfy d X (x, y) ≤ η. Since X is compact and dim(X ) ≤ n, it follows from Proposition 1.4.4 that we can find a finite open cover α = {U0 , U1 , . . . , Ur } of X such that ord(α) ≤ n and mesh(α) < min(ε, η). Choose, for each i ∈ {0, 1, . . . , r }, a point ai ∈ Ui and let pi := f (ai ). By virtue of Lemma 4.7.2, we can find points q0 , q1 , . . . , qr ∈ Rm satisfying the following conditions (C1) the points q0 , q1 , . . . , qr ∈ Rm are in general position, (C2) d( pi , qi ) ≤ 2δ for all 0 ≤ i ≤ r . Let (λi )0≤i≤r be a partition of unity subordinate to the cover α. Let us show that the continuous map g : X → Rm that is defined by g(x) :=

r i=0

has the required properties.

λi (x)qi

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4 Dimension and Maps

Suppose that x and y are points in X such that g(x) = g(y). Denote by s the order at x of α and let us write {i ∈ {0, 1, . . . , r } | x ∈ Ui } = {i 0 , i 1 , . . . , i s }. Similarly, denote by s the order at y of α and let us write { j ∈ {0, 1, . . . , r } | y ∈ U j } = { j0 , j1 , . . . , js }. As s ≤ n ≤ m, Condition (C1) implies that the points qi0 , qi1 , . . . , qis generate an affine subspace A ⊂ Rm of dimension s. Similarly, the points q j0 , q j1 , . . . , q js generate an affine subspace A ⊂ Rm of dimension s . Now observe that A and A both contain the point g(x) = g(y). As a consequence the s + s + 2 points qi0 , qi1 , . . . , qis , q j0 , q j1 , . . . , q js generate an affine subspace of Rm of dimension at most s + s . Since s + s + 2 ≤ 2n + 2 ≤ m + 1, Condition (C1) implies that there exist integers k and k with 0 ≤ k ≤ s and 0 ≤ k ≤ s such that qik = q jk . Therefore, there exists an open subset Ui in the cover α such that x and y both belong to Ui . As diam(Ui ) ≤ mesh(α) < ε, it follows that d(x, y) < ε. Consequently, the map g is ε-injective. It remains to show that d∞ ( f, g) ≤ δ. Let x be an arbitrary point in X . If x ∈ Ui , then we have that d X (x, ai ) ≤ diam(Ui ) ≤ η and hence d( f (x), pi ) = d( f (x), f (ai )) ≤ δ/2. This implies d( f (x),

r

λi (x) pi ) ≤

i=0

≤

r i=0 r i=0

=

λi (x)d( f (x), pi ) λi (x)

δ 2

δ . 2

By the triangle inequality, this gives us

r δ d( f (x), g(x)) ≤ + d λi (x) pi , g(x) 2 i=0

r r δ = +d λi (x) pi , λi (x)qi 2 i=0

≤

δ + 2

r i=0

i=0

λi (x)d( pi , qi )

4.7 Euclidean Embeddings of Finite-Dimensional Spaces

δ δ + λi (x) 2 2

83

r

≤

(by Condition (C2))

i=0

=

δ δ + = δ. 2 2

We deduce that d∞ ( f, g) ≤ δ. This shows that Cε (X, Rm ) is dense in C(X, Rm ). Theorem 4.7.5 Let m and n be non-negative integers such that m ≥ 2n +1. Suppose that X is a compact metrizable space such that dim(X ) ≤ n. Then the set consisting of all continuous maps f : X → Rm that induce a homeomorphism from X onto f (X ) is a G δ dense subset of C(X, Rm ). Proof Choose a metric on the space X that is compatible with its topology. Denote by the subset of C(X, Rm ) consisting of all continuous maps f : X → Rm that induce a homeomorphism from X onto f (X ). Then we clearly have =

∞ k=1

C 1 (X, Rm ). k

As Cε (X, Rm ) is a dense open subset of C(X, Rm ) for every ε > 0 by Lemmas 4.7.3 and 4.7.4, we deduce that is a G δ dense subset of C(X, Rm ) by applying Baire’s theorem (the metric space C(X, Rm ) is complete since Rm is complete). One says that a topological space X embeds in a topological space Y if there exists a subset of Y that is homeomorphic to X . This amounts to saying that there exists a continuous map f : X → Y that induces a homeomorphism from X onto f (X ). Such a map f is called a topological embedding of X in Y . As an immediate consequence of Theorem 4.7.5, we obtain the following result. Corollary 4.7.6 (The Menger-Nöbeling embedding theorem) Every compact metrizable space X satisfying dim(X ) = n embeds in R2n+1 .

Notes The Tietze extension theorem (cf. Theorem 4.1.4), also called the Tietze-Urysohn extension theorem, was first proved for metric spaces by Tietze [104] and then generalized to normal spaces by Urysohn [109]. The notion of a nerve was introduced by Alexandroff in [3]. Theorem 4.4.7 as well as Theorem 4.5.4 are also contained in that paper of Alexandroff. A variant of dimε (X, d) introduced by Gromov [44, Sect. I.1] is Widimε (X, d) which is defined, for any compact metric space (X, d) and ε > 0, as being the smallest integer n such that there exists an ε-injective continuous map f : X → P

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4 Dimension and Maps

from X into some n-dimensional polyhedron P (cf. Exercise 4.11). Motivated by a question raised by Gromov [44, p. 334], Gournay [40] and Tsukamoto [107] obtained interesting estimates for Widimε (X, d) when X is the p -ball in Rn and d is the metric induced by the q -norm. Let X and Y be topological spaces that are not both empty. For X and Y compact and metrizable, it may happen that the inequality dim(X × Y ) ≤ dim(X ) + dim(Y ) in Corollary 4.5.6 is strict. Indeed, in 1930, Pontryagin [91] (see also his survey paper [92, Sect. 11]) gave examples of compact metrizable spaces X and Y with dim(X ) = dim(Y ) = 2 but dim(X × Y ) = 3. Actually, the dimension of the product of two compact metrizable spaces can deviate arbitrarily from the sum of the dimension. More precisely, it was proved in the 1980 s by Dranishnikov (see the survey paper [31]) that, given any positive integers n, m, k with max(n, m)+1 ≤ k ≤ n +m, there exist compact metrizable spaces X and Y satisfying dim(X ) = n, dim(Y ) = m, and dim(X × Y ) = k. Recall that we always have dim(X × Y ) = dim(X ) + dim(Y ) if X and Y are polyhedra by Corollary 3.5.10. The inequality dim(X × Y ) ≤ dim(X ) + dim(Y ) remains valid when X and Y are both compact Hausdorff or both metrizable (see [77], [33, Th. 3.4.9]). In [77], Morita proved the inequality dim(X × Y ) ≤ dim(X )+dim(Y ) in the case when X and Y are paracompact Hausdorff spaces with Y locally compact (see [79, p. 153]). Recall that every metrizable space is paracompact. By a result of Hurewicz [49], one has dim(X × Y ) = dim(X ) + dim(Y ) whenever X is a non-empty compact metrizable space and Y a separable metrizable space with dim(Y ) = 1. In this last result, the compactness hypothesis on X cannot be removed. Indeed, Erdös [34] gave an example of a separable metrizable space X such that dim(X × X ) = dim(X ) = 1 (see Sect. 5.1). On the other hand, Wage [114] described a separable metrizable space X and a paracompact Hausdorff space Y such that dim(X × Y ) = 1 > dim(X ) + dim(Y ) = 0. The result of Corollary 4.7.6 (Menger-Nöbeling theorem) is optimal in the sense that for every integer n ≥ 0 there exists a compact metrizable space X with dim(X ) = n that cannot be embedded in R2n . One can take as X the n-skeleton, i.e., the union of the n-dimensional faces, of a (2n + 2)-simplex (see [33, p. 101]). The idea of using Baire’s theorem in order to prove the Menger-Nöbeling embedding theorem is due to Hurewicz.

Exercises 4.1 Let X be a metric space. Let A and B be disjoint closed subsets of X . Show that the map f : X → [0, 1] defined by f (x) :=

dist(x, A) dist(x, A) + dist(x, B)

is continuous and satisfies A = f −1 (0) and B = f −1 (1).

Exercises

85

4.2 Show that in the statement of Lemma 4.1.2 one cannot replace the condition A ⊂ f −1 (0) by the condition A = f −1 (0). Hint: consider for example the product space X = [0, 1]R with A = {a} and B = {b}, where a and b are distinct points in X . 4.3 Let X be a topological space. Suppose that for every finite open cover α of X , there exists a partition of unity subordinate to α. Show that X is normal. 4.4 Let α = (Ui )i∈I be a finite open cover of a topological space X . Let C denote the geometric realization of the nerve of α. Suppose that ( f i )i∈I and (gi )i∈I are partitions of unity subordinate to α. Let t ∈ [0, 1]. Show that ((1−t) fi +tgi )i∈I is a partition of unity subordinate to α. Deduce that the maps f : X → |C| and g : X → |C| associated with ( f i )i∈I and (gi )i∈I respectively are homotopic, i.e., there exists a continuous map H : X × [0, 1] → |C| such that H (x, 0) = f (x) and H (x, 1) = g(x) for all x ∈ X . 4.5 Let X and Y be compact metrizable spaces. Show that if dim(Y ) = 0, then one has dim(X × Y ) = dim(X ). 4.6 Let X be a non-empty compact metric space. Show that one has dimε (X ) = 0 for every ε > diam(X ). 4.7 Let (X, d X ) and (Y, dY ) be metric spaces. The set X × Y is equipped with the metric d defined by d((x1 , y1 ), (x2 , y2 )) := max(d X (x1 , x2 ), dY (y1 , y2 )) for all (x1 , y1 ), (x2 , y2 ) ∈ X × Y . Show that one has dimε (X × Y, d) ≤ dimε (X, d X ) + dimε (Y, dY ) for every ε > 0. 4.8 Let d1 and d2 be two metrics on a set X . Consider the metric d on X defined by d := max(d1 , d2 ). Show that one has dimε (X, d) ≤ dimε (X, d1 ) + dimε (X, d2 ) for every ε > 0. 4.9 Let d denote the Euclidean metric on [0, 1]2 . Compute dimε ([0, 1]2 , d) for every ε > 0. 4.10 Let n ∈ N and p ∈ [1, ∞]. Denote by d the metric associated with the p-norm · p on Rn . Show that one has dimε (Rn , d) = n for every ε > 0. 4.11 Let (X, d) be a compact metric space and let ε > 0.

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4 Dimension and Maps

(a) Show that there exist a polyhedron P and an ε-injective continuous map f : X → P. (b) Let Widimε (X, d) denote the smallest integer n such that there exist a polyhedron P with dim(P) = n and an ε-injective continuous map f : X → P. Show that one has dimε (X, d) ≤ Widimε (X, d) ≤ 2 dimε (X, d) + 1. (c) Determine dimε (X, d) and Widimε (X, d) for every ε > 0 when X is the Cantor ternary set and d is the usual metric on X ⊂ R.

Chapter 5

Some Classical Counterexamples

The topological spaces presented in this chapter are spaces with amazing properties. Their analysis reveals the validity limits of certain statements in dimension theory and they may be used as counterexamples to various plausible-sounding conjectures. Despite their pathological nature, each of them has its strange intrinsic beauty. In Sect. 5.1 and Sect. 5.2, we construct totally disconnected separable metrizable spaces with positive topological dimension. The space described in Sect. 5.1 is a separable metrizable space that is totally separated but not scattered while the space of Sect. 5.2 is a separable metrizable space that is totally disconnected but not totally separated. Moreover, the space of Sect. 5.1 is a subset of Hilbert space and the space of Sect. 5.2 is obtained by removing a single point from a connected subset of the Euclidean plane. In Sect. 5.3, we construct a countable Hausdorff space with positive topological dimension. The space described in Sect. 5.4 is a zero-dimensional compact Hausdorff space containing an open subset with positive topological dimension. In the last section, we give an example of a topological space with positive topological dimension that is the product of two zero-dimensional normal Hausdorff spaces. None of the results of the present chapter will be used in the sequel.

5.1 The Erdös Space In this section, we describe an example of a separable metrizable space that is totally separated but not scattered. Note that such a space is necessarily totally disconnected with positive topological dimension by Corollaries 2.6.5 and 2.3.3. Let H denote the vector space over R consisting of all real sequences h = (h n )n≥1 that are square-summable, i.e., such that ∞

h 2n < ∞.

n=1

© Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_5

87

88

5 Some Classical Counterexamples

The vector space H is a Hilbert space for the scalar product ·, · defined by h, k =

∞

h n kn

n=1

for all h = (h n ), k = (kn ) ∈ H . We denote by d the associated metric on H . It is given by d(h, k) = h − k, where h =

h, h

is the norm on H associated with the scalar product. Let X denote the subset of H consisting of all sequences x = (xn ) ∈ H such that xn ∈ Q for all n. The space X is called the Erdös space. Proposition 5.1.1 The Erdös space X is separable, metrizable, and totally separated. Proof Of course, X is metrizable since its topology is defined by the metric induced by d on X . The subset Y ⊂ X formed by sequences (xn ) for which xn = 0 for all but finitely many n is countable and dense in X (it is even dense in H ). Therefore, X is separable. Suppose that a = (an ) and b = (bn ) are distinct points in X . Then there exists an integer n 0 such that an 0 = bn 0 . As Q is totally separated, we can find a partition of Q into two disjoint open subsets U and V such that an 0 ∈ U and bn 0 ∈ V (we can take for example U := (ξ, η) ∩ Q and V := Q\U , where ξ and η are irrational / (ξ, η)). Let π : X → Q denote the map numbers such that ξ < an 0 < η and bn 0 ∈ defined by π(x) = xn 0 for all x = (xn ) ∈ X . Observe that π is 1-Lipschitz and hence continuous. Then the open subsets π −1 (U ) and π −1 (V ) form a partition of X and contain respectively a and b. Consequently, the space X is totally separated. Before proving that X is not scattered, let us first establish some auxiliary results. We start with the following elementary observation. Lemma 5.1.2 Let A and B be non-empty disjoint subsets of Q such that Q = A ∪ B. Then, for every ε > 0, there exists a ∈ A and b ∈ B such that |a − b| ≤ ε. Proof Let us inductively construct a sequence (an )n∈N of points of A and a sequence (bn )n∈N of points of B in the following way. We first choose arbitrarily points a0 ∈ A and b0 ∈ B. Suppose now that an ∈ A and bn ∈ B have already been defined for some n ≥ 0. Let cn := (an + bn )/2 ∈ Q denote the middle of the segment [an , bn ]. If cn ∈ A, we take an+1 = cn and bn+1 = bn . Otherwise, we have that cn ∈ B and we take an+1 = an and bn+1 = cn . It then follows that |an − bn | = |a0 − b0 |/2n for all n. Consequently, given ε > 0, we have |an − bn | ≤ ε for n large enough.

5.1 The Erdös Space

89

Proposition 5.1.3 The only bounded clopen subset of X is the empty set. Proof Let be a non-empty bounded open subset of X . Let us show that is not closed in X . Choose a sequence ω = (ω1 , ω2 , . . .) ∈ . Let us construct a sequence of rational numbers r1 , r2 , . . . satisfying, for every integer j ≥ 1, the following conditions: (C1) the sequence u ( j) , defined by u ( j) := (r1 , r2 , . . . , r j , ω j+1 , ω j+2 , . . .), belongs to ; (C2) one has dist(u ( j) , X \) ≤

1 . j

We proceed by induction. Suppose that for some integer i ≥ 0 the rational numbers r 1 , r 2 , . . . , ri ∈ Q satisfying (C1) and (C2), for all j ≤ i, have already been constructed. Consider the map σ: Q → X defined by σ(t) = (r1 , r2 , . . . , ri , t, ωi+2 , ωi+3 , . . .). Note that σ is an isometric embedding of Q into X , that is, d(σ(t), σ(t )) = |t − t | for all t, t ∈ Q. We have that σ(ωi+1 ) = u (i) ∈ since (C1) is satisfied for j = i. On the other hand, the fact that is bounded implies that σ(t) ∈ / for |t| large enough. By applying Lemma 5.1.2 with A = σ −1 () and B = σ −1 (X \), we deduce that we can find a, b ∈ Q such that σ(a) ∈ , σ(b) ∈ X \ and d(σ(a), σ(b)) = |a − b| ≤ 1/(i + 1). Consequently, we can take ri+1 = a. This completes our induction. Consider now the sequence r = (rn )n≥1 . We have that r ∈ X . Indeed, since is bounded, there exists a constant M ≥ 0 such that u ( j) ≤ M for all j ≥ 1. This implies r12 + r22 + · · · + r 2j ≤ M 2 for all j ≥ 1, so that we get, by letting j tend to infinity, ∞ n=1

rn2 ≤ M 2 < ∞.

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5 Some Classical Counterexamples

On the other hand, we have that ∞ ( j) d(r, u ) = (rn − ωn )2 . n= j+1

As r − ω ∈ X , the series

(rn − ωn )2 converges. It follows that d(r, u ( j) ) → 0 as j → ∞,

which shows that r belongs to the closure of in X . Now, by using (C2), we obtain dist(r, X \) = lim j→∞ d(u ( j) , X \) = 0. As X \ is closed in X by hypothesis, we deduce that r ∈ / . This shows that is not closed in X . Corollary 5.1.4 The Erdös space X is not scattered. Proof Let x ∈ X and r > 0. It follows from Proposition 5.1.3 that the open ball B(x, r ) ⊂ X contains no clopen neighborhood of x in X . This shows that X is not scattered.

5.2 The Knaster-Kuratowski Fan In this section, we give an example of a separable metrizable space X that is totally disconnected but not totally separated. Such a space X is not scattered and has positive topological dimension by Proposition 2.6.6 and Corollary 2.3.3. The space X is obtained by removing a point from the Knaster-Kuratowski fan, which is a connected subset of R2 . The construction goes as follows. Consider the Cantor ternary set K ⊂ [0, 1] (see Sect. 2.1). For each c ∈ K , we denote by L c the line segment in the Euclidean plane R2 whose endpoints are (c, 0) and y0 := (1/2, 1/2). We denote by E the subset of K consisting of all the endpoints of the open intervals that are removed from the unit segment [0, 1] in the construction of the Cantor set. In other words, E is the set of all the ternary rational numbers that are in K \{0, 1}: E=

1 2 1 2 7 8 1 2 7 8 19 20 25 26 , , , , , , , , , , , , , ,... . 3 3 9 9 9 9 27 27 27 27 27 27 27 27

Let F := K \E denote the complement of E in K . For each c ∈ E (resp. c ∈ F), we denote by Yc the set consisting of all points in the line segment L c whose ordinate is rational (resp. irrational). The set Y :=

c∈K

Yc ⊂ R2

5.2 The Knaster-Kuratowski Fan

91

Fig. 5.1 The Knaster-Kuratowski fan

is called the Knaster-Kuratowski fan (see Fig. 5.1). Observe that y0 ∈ Y since y0 ∈ Yc for all c ∈ E. The set X := Y \{y0 } is called the punctured Knaster-Kuratowski fan. Proposition 5.2.1 The punctured Knaster-Kuratowski fan X is a totally disconnected separable metrizable space. Proof The space X is separable and metrizable since it is a subset of R2 . Let A be a non-empty connected subset of X . Consider the map π : X → K sending each point x ∈ X to the unique c ∈ K such that x ∈ L c . Observe that π is continuous. As K is totally disconnected, we deduce that π(A) is reduced to a single point c0 ∈ K . We then have A ⊂ Yc0 . As Yc0 is totally disconnected, we conclude that A is also reduced to a single point. This shows that X is totally disconnected. Proposition 5.2.2 The Knaster-Kuratowski fan Y is connected. Proof Let U and V be disjoint open subsets of Y such that U ∪ V = Y and y0 ∈ U . To prove that Y is connected, it suffices to show that U = Y . As the sets U and V are closed in Y , there exists closed subsets A and B of R2 such that U = Y ∩ A and V = Y ∩ B. For each ρ ∈ Q, let us denote by Hρ the horizontal line of R2 consisting of all points whose ordinate is ρ. Let Fρ denote the set consisting of all c ∈ K such that the line segment L c meets Hρ ∩ A ∩ B. Clearly Fρ is closed in K for every ρ ∈ Q. On the other hand, we have that Fρ ⊂ F for every ρ ∈ Q since A ∩ B does not meet Y . The set F has empty interior in K since E is dense in K . As E and Q are countable, it follows that the set M ⊂ K , defined by ⎛ ⎞

M := E ∪ ⎝ Fρ ⎠ , ρ∈Q

is a countable union of closed subsets of K that have empty interior in K . By applying Baire’s theorem, we deduce that the set ⎞ ⎛

Fρ ⎠ K \M = F \ ⎝ ρ∈Q

is dense in K .

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5 Some Classical Counterexamples

Suppose now that c0 ∈ F is such that Yc0 meets B. Denote by t0 the least upper bound of the set consisting of all t ∈ [0, 1/2] such that the point of L c0 with ordinate t belongs to B. Observe that the point in L c0 with ordinate t0 is in A ∩ B. As A ∩ B does not meet Y , it follows that t0 ∈ Q and hence c0 ∈ Ft0 ⊂ M. We deduce that Yc ⊂ A for all c ∈ K \M. As K \M is dense in K , we finally get Y ⊂ A and hence Y = U . This shows that Y is connected. Proposition 5.2.3 The punctured Knaster-Kuratowski fan X is not totally separated. Proof Let c ∈ K . Let x0 and x1 be points in L c ∩ X with ordinate respectively t0 and t1 , where t0 < t1 . Let us show that x1 belongs to the quasi-component of x0 in X . Suppose that it does not. Then it exists a clopen subset U of X such that x0 ∈ U and x1 ∈ X \U . As X \U is open in X , we can find a small Euclidean open disc D ⊂ R2 centered at x1 that does not meet U . Let a and b be elements in R\K such that a < c < b. Denote by the open Euclidean triangle in R2 whose vertices are the points (a, 0), (b, 0) and y0 = (1/2, 1/2). Take a and b sufficiently close to c so that \D has two connected components in R2 , one, denoted by C− , contained in the open half-plane H− consisting of the points in R2 with ordinate < t1 , and the other, denoted by C+ , contained in the open halfplane H+ consisting of the points of R2 with ordinate > t1 (see Fig. 5.2). Then the set E := U ∩ C− is clopen in the fan Y . Indeed, we have that E = U ∩ ∩ H− , which shows that E is open in Y (remark that U is open in X and hence open in Y since X is open in Y ). On the other hand, we also have E = U ∩ ∩ H− , where H− denotes the closure of H− in R2 . As U and ∩ X are closed in X , we deduce that E is closed in X . It follows that E is also closed in Y since we can find a small open Euclidean ball centered at y0 that does not meet E. As Y is connected by Proposition 5.2.2 and E = ∅ since x0 ∈ E, we deduce that E = Y . This gives a contradiction because

Fig. 5.2 Proof of Proposition

5.2 The Knaster-Kuratowski Fan

93

x1 ∈ / E. Consequently, the quasi-component of x0 in X contains x1 and hence X is not totally separated.

5.3 The Bing Space In this section, we provide an example of a countably-infinite connected Hausdorff space. Such a space has necessarily positive covering dimension by Proposition 1.3.3. Let X denote the closed upper half-plane in Q2 , i.e., the set consisting of all pairs (x, y) ∈ Q2 such that y ≥ 0. Let us fix some irrational real number θ > 0. We define a topology on X in the following way. For each (x, y) ∈ X , define the numbers p− (x, y) and p+ (x, y) by p− (x, y) := x −

y y and p+ (x, y) := x + . θ θ

Thus, the point ( p− (x, y), 0) (resp. ( p+ (x, y), 0)) is the intersection point of the line with slope θ (resp. −θ) passing through the point (x, y) with √ the horizontal axis R × {0} ⊂ R2 (see Fig. 5.3). Note that, in the case when θ = 3, the points (x, y), ( p− (x, y), 0), and ( p+ (x, y), 0) are the vertices of a Euclidean equilateral triangle. Given a real number ε > 0, we denote by Vε (x, y) the set consisting of the point (x, y) and all the points (z, 0) ∈ X satisfying |z − p− (x, y)| < ε or |z − p+ (x, y)| < ε. Thus, we have Vε (x, y) := {(x, y)} ∪ Iε (x, y) ∪ Jε (x, y), where Iε (x, y) (resp. Jε (x, y)) denotes the set of rational points on the horizontal axis that belong to the open interval of length 2ε centered at ( p− (x, y), 0) (resp. ( p+ (x, y), 0)) (see Fig. 5.3).

Fig. 5.3 Construction of the Bing space

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5 Some Classical Counterexamples

Let T denote the set consisting of all the subsets ⊂ X satisfying the following condition: for every (x, y) ∈ , there exists a rational number ε > 0 such that Vε (x, y) ⊂ . One easily checks that T satisfies: (1) ∅, X ∈ T , (2) T is closed under finite intersections, (3) T is closed under arbitrary unions. In other words, T is the set of open sets for a topology on X . We equip the set X with this topology. The topological space X is called the Bing space. Proposition 5.3.1 The Bing space X is a countably-infinite second-countable Hausdorff space. Proof The set X is countably-infinite because the set Q is. The sets Vε (x, y), where (x, y) ∈ X and ε > 0 is rational, clearly form a countable base for the topology on X . Therefore X is second-countable. Let (x, y) and (x , y ) be distinct points in X . Suppose first that neither (x, y) nor (x , y ) is on the horizontal axis. Then, as θ is irrational, the points p− (x, y), p− (x , y ), p+ (x, y) and p+ (x , y ) are all distinct. In the case when one of the points (x, y) and (x , y ), say (x, y), is on the horizontal axis and the other is not, then we have that p− (x , y ) = p− (x, y) = p+ (x, y) = p+ (x , y ). Finally if both (x, y) and (x , y ) are on the horizontal axis, we have that p− (x, y) = p+ (x, y) = p− (x , y ) = p+ (x , y ). In all cases, we see that the neighborhoods Vε (x, y) and Vε (x , y ) do not meet for ε small enough. This shows that X is Hausdorff. Lemma 5.3.2 Let (x, y) and (x , y ) be two points in X . Then one has Vε (x, y) ∩ Vε (x , y ) = ∅ for all ε > 0 and ε > 0. Proof A point (u, v) ∈ X belongs to the closure of Vε (x, y) if and only if | p− (u, v) − p− (x, y)| ≤ ε or | p− (u, v) − p+ (x, y)| ≤ ε or | p+ (u, v) − p− (x, y)| ≤ ε or | p+ (u, v) − p+ (x, y)| ≤ ε. We deduce that, for ε small enough, Vε (x, y) is the union of four “strips”, two with slope θ and two with slope −θ. From this description, it is clear that Vε (x, y) always meets Vε (x , y ). Alternatively, assuming for instance p− (x, y) ≤ p− (x , y ), a point (u, v) ∈ Vε (x, y) ∩ Vε (x , y ) may be explicitly obtained by solving the system

u− u+

v θ v θ

=a =b

with u, v ∈ Q and a, b ∈ R such that a ≤ b, |a − p− (x, y)| ≤ ε and |b − p+ (x , y )| ≤ ε . Lemma 5.3.3 Let U and U be non-empty open subsets of X . Then one has U ∩ U = ∅.

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95

Proof Let (x, y) ∈ U and (x , y ) ∈ U . As U and U are open subsets, there exist ε > 0 and ε > 0 such that Vε (x, y) ⊂ U and Vε (x , y ) ⊂ U . We have that Vε (x, y) ∩ Vε (x , y ) ⊂ U ∩ U . Since Vε (x, y) ∩ Vε (x , y ) = ∅ by Lemma 5.3.2, we deduce that U ∩ U = ∅. Proposition 5.3.4 The Bing space X is connected. Proof If U is a clopen subset of X then U := X \U is also clopen so that U ∩ U = U ∩ U = ∅. This implies U = ∅ or U = ∅ by Lemma 5.3.3. Therefore X is connected. Corollary 5.3.5 The Bing space X has topological dimension dim(X ) ≥ 1. Proof By Proposition 1.3.3, every connected accessible space X having more than one point satisfies dim(X ) ≥ 1. Corollary 5.3.6 The Bing space X is not normal and hence not metrizable. Proof By Corollary 1.7.2, every non-empty countable normal space N satisfies dim(N ) = 0.

5.4 The Tychonoff Plank In this section, we give an example of a scattered locally compact Hausdorff space with positive covering dimension. Let X be a set endowed with a total ordering denoted by ≤. The open intervals in X , that is, X itself and all the intervals of the form (←, x) := {z ∈ X | z < x}, (x, →) := {z ∈ X | x < z}, and (x, y) := {z ∈ X | x < z < y}, where x and y run over X , constitute a base for a topology on X . This topology is called the order topology associated with the ordering ≤. Proposition 5.4.1 Let X be a totally ordered set endowed with the associated order topology. Then X is Hausdorff. Proof Let x and y be distinct points in X . Let us show that there exist an open neighborhood U of x and an open neighborhood V of y such that U ∩ V = ∅. We can assume x < y. If the interval (x, y) is empty, we can take U := (←, y) and V := (x, →). Otherwise, there exists z ∈ X such that x < z < y. We can then take U := (←, z) and V := (z, →). This shows that X is Hausdorff. Let A be a subset of a totally ordered set X . Recall that one says that an element x ∈ X is a lower bound (resp. an upper bound) for A if we have that x ≤ a (resp. a ≤ x) for all a ∈ A. One says that x is a greatest lower bound (resp. a least upper bound) for A if x is a lower bound (resp. an upper bound) for A and m ≤ x (resp. x ≤ m) for every lower bound (resp. every upper bound) m for A. Note that if A admits a greatest lower bound (resp. a least upper bound) then it is unique.

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Proposition 5.4.2 Let X be a totally ordered set endowed with the associated order topology. Then X is compact if and only if every non-empty subset of X admits both a greatest lower bound and a least upper bound. Proof Let us first show that this condition is necessary. Suppose that X is compact and let A be a non-empty subset of X . Then the set A admits a greatest lower bound since otherwise the intervals (a, →) and (←, m), where a runs over A and m runs over all lower bounds of A, would form an open cover of X admitting no finite subcover. A similar argument shows that A admits a least upper bound. Conversely, suppose that X = ∅ and that every non-empty subset of X admits a greatest lower bound and a least upper bound. This implies in particular that X admits a minimal element m 0 and a maximal element M0 . Let α be an open cover of X . Consider the subset A of X consisting of all x ∈ X such that the half-open interval [m 0 , x) can be covered by a finite number of elements of α. Observe that A is not empty since m 0 ∈ A. Let M denote the least upper bound of A. As α covers X , we can find an open set U in the family α such that M ∈ U . We claim that M = M0 . Indeed, otherwise, there would exist elements y, z ∈ X such that y < M < z and (y, z) ⊂ U . This would imply z ∈ A and would contradict the fact that M is an upper bound for A. This proves that M = M0 . It follows that α admits a finite subcover. Therefore, the space X is compact. A well-ordered set is a set X equipped with an ordering relation such that every non-empty subset of X has a minimal element. The ordering relation of a wellordered set X is a total ordering since the set {x, y} has a minimal element for all x, y ∈ X . Corollary 5.4.3 Let X be a non-empty well-ordered set endowed with the associated order topology. Then the space X is compact if and only if X has a maximal element. Proof This is a necessary condition. Indeed, if X is compact, it follows from Proposition 5.4.2 that X has a least upper bound. This least upper bound is the maximal element of X . Let us show now that this condition is also sufficient. Suppose that X admits a maximal element M0 . Let A be a non-empty subset of X . As X is well-ordered, A has a minimal element and hence a greatest lower bound. On the other hand, since X is well-ordered, the set E consisting of all upper bounds of A, which is not empty as M0 ∈ E, has a minimal element. Therefore, the set A admits a least upper bound. We deduce that X is compact by applying Proposition 5.4.2. Let us recall without proofs some basic facts about ordinal numbers (see for example [58] for more details). Two well-ordered sets E and F are called isomorphic if there exists an orderpreserving bijective map from E onto F. An ordinal is an isomorphism class of well-ordered sets. If E is a well-ordered set that represents an ordinal ξ, one also says that ξ is the order type of E. There is no set containing all ordinals (this would lead to a contradiction in set theory). One speaks instead of the collection of all ordinals. Let ξ and η be ordinals,

5.4 The Tychonoff Plank

97

represented by well-ordered sets E and F respectively. We write ξ ≤ η if E is isomorphic to a subset of F. It can be shown that the relation ≤ is a well-ordering on the collection of all ordinals. For each n ∈ N, we denote by n the ordinal represented by the well-ordered set {1, . . . , n}. Let ξ be an ordinal. The interval [0, ξ), which consists of all ordinals α < ξ, is a well-ordered set whose order type is ξ. The minimal element of the interval (ξ, →) is the order type of the well-ordered set [0, ξ]. This ordinal is denoted by ξ + 1 and is called the successor of ξ. Observe that, for every ordinal α < ξ, we have (α, ξ] = (α, ξ + 1) = [α + 1, ξ].

(5.4.1)

Proposition 5.4.4 Let ξ be an ordinal. Then the set X := [0, ξ], equipped with its order topology, is a scattered compact Hausdorff space. Proof The space X is Hausdorff by Proposition 5.4.1. The compactness of X follows from Corollary 5.4.3. To prove that X is scattered, it suffices to verify that every point η ∈ X admits a neighborhood base consisting of clopen subsets of X . We can assume 0 < η ≤ ξ since 0 is isolated in X . Then the intervals of the form (α, η], where α < η, form a neighborhood base of η. These intervals are clopen in X by (5.4.1). Consequently, the space X is scattered. An ordinal ξ is said to be infinite (resp. countable) if the well-ordered sets that represent ξ are infinite (resp. countable). Note that there exist uncountable ordinals since every set can be well-ordered. We denote by ω the smallest infinite ordinal (i.e., the countable ordinal that is the order type of N) and we denote by the smallest uncountable ordinal. Let us equip the sets [0, ω] and [0, ] with their order topology. The product space P := [0, ] × [0, ω] is called the Tychonoff plank (Fig. 5.4). Proposition 5.4.5 The Tychonoff plank P is a scattered compact Hausdorff space. Proof The spaces [0, ] and [0, ω] are scattered compact Hausdorff spaces by Proposition 5.4.4 and any product of Hausdorff (resp. compact, resp. scattered) spaces is Hausdorff (resp. compact, resp. scattered). Corollary 5.4.6 The Tychonoff plank P has topological dimension dim(P) = 0. Proof The space P is compact and hence Lindelöf. On the other hand, by Theorem 2.4.20, every non-empty scattered Lindelöf space L satisfies dim(L) = 0.

Fig. 5.4 The Tychonoff plank

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5 Some Classical Counterexamples

The space X := P\ (, ω) obtained from the Tychonoff plank by removing its right upper point, is called the punctured Tychonoff plank. Proposition 5.4.7 The punctured Tychonoff plank X is a scattered locally compact Hausdorff space. Proof The space P is a scattered compact Hausdorff space by Proposition 5.4.5. On the other hand, every subspace of a Hausdorff (resp. scattered) space is itself Hausdorff (resp. scattered) and every open subspace of a compact Hausdorff space is locally compact. Proposition 5.4.8 The punctured Tychonoff plank X is not normal. Proof Let A := {} × [0, ω) and B := [0, ) × {ω}. Thus, A is the subset of X consisting of all points whose first coordinate is while B is the subset of X consisting of all points whose second coordinate is ω. The sets A and B are closed subsets of X since A = X ∩ ({} × [0, ω]) and B = X ∩ ([0, ] × {ω}). Observe that A = {(, n) | n ∈ N}. Let U be an open subset of X containing A. As U is a neighborhood of every point in A, there exists, for each n ∈ N, an ordinal ξn < such that (ξn , ] × {n} ⊂ U . Denote by η the least upper bound of the set consisting of all such ξn . We have that [0, η) =

[0, ξn ).

n∈N

As the ordinals ξn are countable, we deduce that η is itself countable. It follows that η < . As (η, ] × [0, ω) ⊂ U , we conclude that every neighborhood of the point (η + 1, ω) ∈ B meets U . Consequently, every open subset of X containing B meets U . As the sets A and B are disjoint closed subsets of X , this shows that the space X is not normal. Note however that P is normal since it is compact and Hausdorff (Proposition 1.5.4). This shows that a subspace of a normal space may fail to be normal. Corollary 5.4.9 The punctured Tychonoff plank X has topological dimension dim(X ) ≥ 1. Proof Every topological space whose covering dimension is 0 is normal by Corollary 2.3.2. Remark 5.4.10 Recall that the Tychonoff plank P has covering dimension dim(P) = 0 by Corollary 5.4.6. Thus, Corollary 5.4.9 shows that a subspace of a topological space with covering dimension 0 may have positive covering dimension.

5.4 The Tychonoff Plank

99

Remark 5.4.11 The space P is a Lindelöf space since it is compact. However, X is not a Lindelöf space since every non-empty scattered Lindelöf space L satisfies dim(L) = 0 by Theorem 2.4.20. This shows that a subspace of a Lindelöf space is not necessarily Lindelöf. Corollary 5.4.12 The Tychonoff plank P and the punctured Tychonoff plank X are not metrizable. Proof The punctured Tychonoff plank is not metrizable since every metrizable space is normal by Proposition 1.5.3. As every subspace of a metrizable space is metrizable, the Tychonoff plank is not metrizable either.

5.5 The Sorgenfrey Plane This section is devoted to the Sorgenfrey plane, a topological space that may be used to show that the product of two normal (resp. Lindelöf, resp. zero-dimensional) spaces may fail to be normal (resp. Lindelöf, resp. zero-dimensional). We start by introducing the Sorgenfrey line. Consider the set B consisting of all half-open intervals of R that are of the form [a, b), where a and b run over R. It is clear that B covers R and that B1 ∩ B2 ∈ B for all B1 , B2 ∈ B. Thus, there is a unique topology on R admitting B as a base. The Sorgenfrey line is the topological space S with underlying set R that admits B as a base. A subset ⊂ S is open if and only if it satisfies the following condition: for every x ∈ , there exists ε > 0 such that [x, x + ε) ⊂ . It follows in particular that the topology on S is finer than the usual topology on R. In fact, it is strictly finer since for example the half-open intervals [a, b), where a < b, are open in S while they are not open for the usual topology on R. Note that [a, b) is also closed in S since S\[a, b) = (−∞, a) ∪ [b, ∞) is clearly open in S. Proposition 5.5.1 The Sorgenfrey line S is a scattered Lindelöf first-countable separable Hausdorff space. Proof The topological space S is scattered since the half-open intervals [a, b) are clopen in S and form a base of the topology. The topology on S is Hausdorff since it is finer than the usual topology on R. For every x ∈ S, the intervals [x, q), where q ∈ Q and x < q, form a countable neighborhood base of x. Therefore S is first-countable. The set Q is dense in S since every interval [a, b), where a < b, contains rational numbers. Therefore S is separable. It remains only to show that S is Lindelöf. Suppose that α = (Ui )i∈I is an open cover of S. Consider the subset X ⊂ S consisting of all points x ∈ S satisfying the following condition: there exist two real numbers ax < bx and an element i(x) ∈ I such that x ∈ (ax , bx ) ⊂ Ui(x) . Observe now that the family (X ∩ (ax , bx ))x∈X is an open cover of X with respect to the usual topology. As X ⊂ R with its usual topology

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5 Some Classical Counterexamples

is Lindelöf by Corollary 2.4.14, it follows that there exists a countable subset C ⊂ X such that the subfamily (X ∩ (ax , bx ))x∈C covers X . This implies X ⊂ j∈J U j , where J := {i(x) | x ∈ C} is countable. On the other hand, we claim that Y := S\X is countable. Indeed, the fact that α is an open cover of S implies that, for every y ∈ Y , there exist real numbers a y < b y and an element i(y) ∈ I such that y ∈ [a y , b y ) ⊂ Ui(y) . Then we can find a rational number ρ(y) ∈ [a y , b y ). If y1 , y2 ∈ Y satisfy y1 < y2 , then ρ(y1 ) = ρ(y2 ) since otherwise we would get y2 ∈ X . Therefore the map ρ : Y → Q is injective. We deduce that Y is countable. Therefore, we can find a countable subset K ⊂ I such that Y ⊂ k∈K Uk . We conclude that (Ui )i∈J ∪K is a countable cover of X . This shows that X is Lindelöf. Corollary 5.5.2 The Sorgenfrey line S has topological dimension dim(S) = 0. Proof By Theorem 2.4.20, every non-empty scattered Lindelöf space X has covering dimension dim(X ) = 0. Corollary 5.5.3 The Sorgenfrey line S is normal. Proof By Corollary 2.3.2, every topological space X satisfying dim(X ) = 0 is normal. The Sorgenfrey plane is the space S × S, i.e., the product of the Sorgenfrey line with itself. Proposition 5.5.4 The Sorgenfrey plane S × S is a scattered first-countable separable Hausdorff space. Proof This immediately follows from Proposition 5.5.1 since the product of two scattered (resp. first-countable, resp. separable, resp. Hausdorff) spaces is itself scattered (resp. first-countable, resp. separable, resp. Hausdorff). The following observation will be useful. Lemma 5.5.5 The second diagonal := {(x, −x) | x ∈ S} is discrete and closed in the Sorgenfrey plane S × S. Proof For every p = (x, y) ∈ S × S and ε > 0, the set C( p, ε) := [x, x + ε) × [y, y + ε) is open in S × S. If p lies outside of and ε is small enough, then C( p, ε) does not meet . Consequently, is closed in S × S. On the other hand, is discrete in S × S since ∩ C( p, ε) = { p} for all p ∈ and ε > 0. Proposition 5.5.6 The Sorgenfrey plane S × S is not Lindelöf. Proof The second diagonal := {(x, −x) | x ∈ S} is a closed discrete subset of S × S by Lemma 5.5.5. As every discrete Lindelöf space is countable (see Example 2.4.5), we deduce that is not Lindelöf. It follows that S × S is not Lindelöf since every closed subset of a Lindelöf space is itself Lindelöf by Proposition 2.4.6.

5.5 The Sorgenfrey Plane

101

Corollary 5.5.7 The Sorgenfrey line S and the Sorgenfrey plane S × S are not second-countable. Proof The Sorgenfrey plane S × S is not second-countable since every secondcountable space is Lindelöf by Proposition 2.4.11. As the product of two secondcountable spaces is also second-countable by Proposition 2.4.10, this implies that the Sorgenfrey line S is not second-countable either. Corollary 5.5.8 The Sorgenfrey line S and the Sorgenfrey plane S × S are not metrizable. Proof The Sorgenfrey plane S × S is not metrizable since every separable metrizable space is Lindelöf by Proposition 2.4.18. As the product of two metrizable spaces is also metrizable, this implies that the Sorgenfrey line S is not metrizable either. Proposition 5.5.9 The Sorgenfrey plane S × S is not normal. Proof First observe that, since S × S is separable by Proposition 5.5.4, the set F consisting of all continuous maps f : S × S → R has cardinality bounded above by the cardinality c of the continuum, i.e., card(F) ≤ card(R) = c. On the other hand, since is discrete, the characteristic map χ A : → R is continuous for every subset A ⊂ . If S × S were normal, it would be possible to extend every χ A to a continuous map f A : S × S → R by applying the Tietze extension theorem (Theorem 4.1.4). This would imply that the cardinality of F is at least that of the power set of , i.e., card(F) ≥ 2c. We would then get a contradiction since, by Cantor’s theorem, we have that 2ξ > ξ for every cardinal ξ. Consequently, the Sorgenfrey plane is not normal. Remark 5.5.10 The Sorgenfrey plane is separable but its second diagonal ⊂ S × S is not since it is discrete and uncountable. This shows that a subspace of a separable space may fail to be separable even if it is closed. Note however that every open subspace of a separable space is clearly separable and that it follows from Propositions 2.4.9 and 2.4.18 that every subspace of a separable metrizable space is separable. Corollary 5.5.11 The Sorgenfrey plane S × S has topological dimension dim(S × S) ≥ 1. Proof Every topological space X satisfying dim(X ) = 0 is normal by Corollary 2.3.2.

Notes The counterexamples gathered in this chapter played an important role in the history of dimension theory (see [33, 50, 93]). They are named after the mathematicians who discovered them.

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5 Some Classical Counterexamples

The space X of Sect. 5.1 was introduced by Erdös in [34]. It has topological dimension dim(X ) = ind(X ) = Ind(X ) = 1 (see [34]) and Roberts [95] proved that it can be embedded in the Euclidean plane R2 . Note that X is a subgroup of the additive group H and hence inherits a structure of a topological group. This gives an interesting example of a totally disconnected abelian group. It turns out that X is isomorphic, as a topological group, to certain homeomorphism groups of manifolds (see [29] and the references therein). The Knaster-Kuratowski fan was described in [60] (see [102, Example 129 p. 145] and [33, p. 29]). It is also sometimes called the Cantor teepee. The point y0 is a dispersion point of the Knaster-Kuratowski fan Y (a point p in a connected space C is called a dispersion point if the space C\{ p} is totally disconnected). Propositions 5.2.3, 2.6.6, and Corollary 2.3.3 imply that the punctured KnasterKuratowski fan X = Y \{y0 } satisfies dim(X ) ≥ 1. As X ⊂ Y ⊂ R2 , we deduce that 1 ≤ dim(X ) ≤ dim(Y ) ≤ 2 by applying Theorem 1.8.3 and Corollary 3.5.7. Actually, it can be shown that dim(X ) = dim(Y ) = 1 by using the fact that every subset of Rn whose topological dimension is n has non-empty interior (see for example [50, Th. IV.3 p. 44]). The counterexample of Sect. 5.3 was described by Bing in a one-page paper [14, Example 1] (see [102, p. 93] and [17, I p. 108 exerc. 21 and I p. 115 exerc. 1]). The Bing space has covering dimension ∞ and small inductive dimension 1 (cf. [14]). The first examples of countably infinite connected Hausdorff spaces were given by Urysohn in his posthumous article [109]. Subsequently, many other interesting examples of such spaces were discovered (see the paper by Miller [75] and the references therein). A topological space X is called a Urysohn space if any two distinct points of X admit disjoint closed neighborhoods. Of course, every Urysohn space is Hausdorff. It immediately follows from Lemma 5.3.3 that the Bing space is not a Urysohn space. An example of a countably-infinite connected Urysohn space admitting a dispersion point was constructed by Roy in [97]. The Tychonoff plank was introduced in [108]. The smallest uncountable ordinal is sometimes denoted ω1 instead of . It can be shown that the punctured Tychonoff plank X has covering dimension dim(X ) = 1. The Sorgenfrey topology was used by Sorgenfrey in [100] to show that a product of paracompact spaces is not necessarily paracompact, thus settling in the negative a question previously raised by Dieudonné [28]. According to Cameron [20], it seems that the copaternity of the Sorgenndroff and Urysohn [10] for priority reasons.

Exercises 5.1 Let X denote the Erdös space (cf. Sect. 5.1). (a) Show that X and QN are isomorphic as vector spaces over Q and hence as additive groups. (b) Describe an injective continuous map f : X → QN .

Exercises

103

(c) Show that the space QN is scattered. (d) Show that there is no subspace of QN that is homeomorphic to X . 5.2 Show that dim(QN ) = 0. 5.3 Show that the Erdös space X described in Sect. 5.1 cannot be embedded into R. 5.4 Show that the space QN can be embedded into the Cantor set (and hence into R). 5.5 Let X denote the Erdös space (cf. Sect. 5.1). Show that X × X is homeomorphic to X . 5.6 In this exercise, we use the notation of Sect. 5.2. Let x be a point in X and let c ∈ K such that x ∈ Yc . Show that the quasi-component of x in X is Yc \{y0 }. 5.7 Show that the Bing space X described in Sect. 5.3 is not regular. (Recall that a topological space X is called regular if for every closed subset A of X and every point x ∈ X \A, there exist disjoint open subsets U and V of X such that A ⊂ U and x ∈ V .) 5.8 Let X be the Bing space described in Sect. 5.3. Consider the subsets Y and Z of X defined by Y := Q × {0} ∼ = Q and Z := X \Y = {(x, y) ∈ X | y > 0}. (a) (b) (c) (d)

Show that the topology induced by X on Z is the discrete one. Show that the topology induced by X on Y is the usual topology on Q. Show that Y is an open dense subset of X . Give a direct proof of the fact that X is not compact by showing that the cover of X consisting of Y and all the subsets of the form Y ∪ {z}, where z ∈ Z , is an open cover admitting no finite subcover.

5.9 Let X denote the Bing space described in Sect. 5.3. Show that X admits a base consisting of open subsets whose topological boundary is clopen. 5.10 Show that a countable connected Hausdorff space having more than one point cannot be compact. 5.11 Show that a countable accessible topological space having more than one point cannot be path-connected. Hint: use Baire’s theorem to prove that the unit segment [0, 1] cannot be expressed as the union of a countably-infinite family of pairwise disjoint non-empty closed subsets. 5.12 Let X be a countable connected space. Show that every continuous map f : X → R is constant. 5.13 (The relatively prime topology on the positive integers [39], [102, Example 60 p. 82]). Let Z+ := {1, 2, . . .} denote the set of positive integers. Given coprime integers x, r ∈ Z+ , define the subset Vr (x) ⊂ Z+ by Vr (x) := {x + r n | n ∈ Z} ∩ Z+ .

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(a) Show that there is a unique topology on Z+ admitting as a base the set consisting of all Vr (x), where x, r ∈ Z+ are coprime. In the sequel, the set Z+ is equipped with this topology. (b) Show that the space Z+ is Hausdorff. (c) Show that if x, r ∈ Z+ are coprime, then every y ∈ Z+ that is a multiple of r belongs to the closure of Vr (x). (d) Deduce from (c) that if U and U are non-empty open subsets of Z+ then one has U ∩ U = ∅. (e) Show that the space Z+ is connected. 5.14 The set [0, ), which consists of all the ordinals that are smaller than the first uncountable ordinal , is equipped with its order topology. Show that [0, ) is a locally compact first-countable Hausdorff space. Show that [0, ) is not Lindelöf. Show that [0, ) is not second-countable. Show that [0, ) is sequentially-compact but not compact. (Recall that a topological space X is called sequentially-compact if every sequence of points of X admits a convergent subsequence.) (e) Show that [0, ) is not metrizable. (f) Recover from (e) the fact that neither the punctured Tychonoff plank P nor the Tychonoff plank X is metrizable (cf. Corollary 5.4.12).

(a) (b) (c) (d)

5.15 Let denote the first uncountable ordinal. Let [0, ] be the set consisting of all ordinals ξ ≤ , equipped with its order topology. (a) Show that [0, ] is not first-countable. (b) Deduce from (a) that neither the Tychonoff plank P nor the Tychonoff plank X is first-countable. 5.16 Show that the punctured Tychonoff plank X is not σ-compact. 5.17 Show that neither the Tychonoff plank P nor the punctured Tychonoff plank X are separable. 5.18 Show that every non-empty subspace X of the Sorgenfrey line S has topological dimension dim(X ) = 0. 5.19 Show that every subspace of the Sorgenfrey line S is normal. 5.20 Let A and B be the subsets of the Sorgenfrey plane S × S defined by A := {(x, −x) ∈ S × S | x ∈ Q} and B := {(x, −x) ∈ S × S | x ∈ / Q}. Give a direct proof that S × S is not normal (cf. Proposition 5.5.9) by showing that there do not exist disjoint open subsets U and V of S × S with A ⊂ U and B ⊂ V.

Part II

Mean Topological Dimension

Chapter 6

Mean Topological Dimension for Continuous Maps

In this chapter, the term “dynamical system” refers to a pair (X, T ), where X is a topological space and T a continuous map from X into itself. The topological space X is called the phase space of the dynamical system and it will be implicitly assumed to be non-empty. The self-mapping T describes the way the points are moved in the phase space when times goes from t to t + 1. At time t = n, the change in the phase space from t = 0 is described by the iterate T n = T ◦ T ◦ · · · ◦ T (n times). In Sect. 6.3, we define the mean topological dimension, denoted by mdim(X, T ), of a dynamical system (X, T ), where X is a normal space and T : X → X a continuous map. This definition is an asymptotic version of the definition of covering dimension for topological spaces that is presented in Sect. 1.1. Mean topological dimension is an invariant of topological conjugacy taking its values in [0, ∞]. We have that mdim(X, T ) = 0 whenever X has finite topological dimension (Proposition 6.4.4). Examples of dynamical systems with positive mean topological dimension will be given in the next chapter. When X is a compact metric space, we give in Sect. 6.5 an equivalent definition of mdim(X, T ) that involves the metric.

6.1 Joins Let α = (Ai )i∈I and β = (B j ) j∈J be two families of subsets of a set X . The join of α and β is the family α ∨ β of subsets of X defined by α ∨ β := (Ai ∩ B j )(i, j)∈I ×J . Proposition 6.1.1 Let f : X → Y be a map from a set X into a set Y . Let α and β be two families of subsets of Y . Then one has f −1 (α ∨ β) = f −1 (α) ∨ f −1 (β).

© Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_6

107

108

6 Mean Topological Dimension for Continuous Maps

Proof Suppose that α = (Ai )i∈I and β = (B j ) j∈J . Then we have f −1 (α ∨ β) = ( f −1 (Ai ∩ B j ))(i, j)∈I ×J = ( f −1 (Ai ) ∩ f −1 (B j ))(i, j)∈I ×J = f −1 (α) ∨ f −1 (β).

Remark 6.1.2 If α and β are covers of X , then α ∨ β is also a cover of X and one has α ∨ β α and α ∨ β β. Moreover, if γ is a cover of X such that γ α and γ β, then one has γ α ∨ β. Remark 6.1.3 If α and β are open (resp. closed) covers of a topological space X , then α ∨ β is an open (resp. closed) cover of X . Lemma 6.1.4 Let X , Y and Z be topological spaces. Let α and β be finite open covers of X . Suppose that f : X → Y is an α-compatible continuous map and that g : X → Z is a β-compatible continuous map. Then the continuous map F : X → Y × Z , defined by F(x) := ( f (x), g(x)) for all x ∈ X , is (α ∨ β)-compatible. Proof Let γ = (Vi )i∈I be a finite open cover of Y such that f −1 (γ) α and let δ = (W j ) j∈J be a finite open cover of Z such that g −1 (δ) β. Then η := (Vi × W j )(i, j)∈I ×J is a finite open cover of Y × Z satisfying F −1 (η) α ∨ β. This shows that F is (α ∨ β)-compatible. Proposition 6.1.5 Let X be a normal space. Let α and β be finite open covers of X . Then one has D(α ∨ β) ≤ D(α) + D(β). Proof By Proposition 4.4.6, we can find a polyhedron P such that dim(P) = D(α) and an α-compatible continuous map f : X → P. Similarly, we can find a polyhedron Q such that dim(Q) = D(β) and a β-compatible continuous map g : X → Q. Then the continuous map F : X → P × Q, defined by F(x) := ( f (x), g(x)) for all x ∈ X , is (α ∨ β)-compatible by Lemma 6.1.4. We deduce that D(α ∨ β) ≤ dim(P × Q) by using Proposition 4.4.5. As dim(P × Q) = dim(P) + dim(Q) by Corollary 3.5.10, we finally get D(α ∨ β) ≤ dim(P) + dim(Q) = D(α) + D(β).

The following example shows that Proposition 6.1.5 becomes false if we drop the normality hypothesis on X . Example 6.1.6 Let X = {x0 , a, b, c, d} be a set of cardinality 5. Equip X with the topology for which the open sets are the empty set and all the subsets of X containing x0 . Note that dim(X ) = 3 (see Example 1.1.11). Consider the open cover α of X that consists of the sets {x0 , a, b} and {x0 , c, d} and the open cover β that consists of the sets {x0 , a, c} and {x0 , b, d}. Then we clearly have D(α) = D(β) = 1. However, the cover α ∨ β consists of the sets {x0 , a}, {x0 , b}, {x0 , c}, and {x0 , d} so that D(α ∨ β) = 3 and hence D(α ∨ β) > D(α) + D(β).

6.2 Subadditive Sequences

109

6.2 Subadditive Sequences A sequence (u n )n≥1 of real numbers is said to be subadditive if it satisfies u n+m ≤ u n + u m for all n, m ≥ 1. √ Example 6.2.1 The sequences (n)n≥1 , ( n)n≥1 , (1 + (−1)n )n≥1 , and (−n 2 )n≥1 are all subadditive. Example 6.2.2 Let a, b, c be non-negative real numbers. If the sequences (u n )n≥1 and (vn )n≥1 are subadditive, then the sequence (a + bu n + cvn )n≥1 is subadditive. Proposition 6.2.3 (Fekete’s lemma) Let (u n )n≥1 be a subadditive sequence of real numbers such that u n ≥ 0 for all n ≥ 1. Then the sequence (vn )n≥1 defined by un is convergent. Moreover, one has vn := n lim vn = inf vn .

n→∞

n≥1

(6.2.1)

Proof Let n and k be integers such that n > k ≥ 1. Then there are unique integers q ≥ 1 and r ∈ {1, . . . , k} such that n = qk + r . The subadditivity of the sequence (u n ) gives us vn =

u qk+r qu k + u r qu k ur ur ≤ ≤ + = vk + , n n qk n n

which implies vn ≤ vk +

max(u 1 , . . . , u k ) . n

By letting n tend to infinity, we deduce that lim sup vn ≤ vk n→∞

for all k ≥ 1. Consequently, we have that lim sup vn ≤ λ := inf vn . n→∞

n≥1

As λ ≤ lim inf vn ≤ lim sup vn , n→∞

it follows from inequality (6.2.2) that

n→∞

(6.2.2)

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6 Mean Topological Dimension for Continuous Maps

lim sup vn = lim inf vn = λ. n→∞

n→∞

This shows that the sequence (vn ) converges to λ.

Example 6.2.4 Let X be a non-empty compact metrizable space. Then X n is also compact and metrizable for any n ≥ 1. As X n embeds as a closed subset of X n+1 = X n × X for any n, the sequence (dim(X n ))n≥1 is non-decreasing. Moreover, if dim(X ) < ∞, it follows from Corollary 4.5.6 that (dim(X n ))n≥1 is a subadditive sequence of non-negative integers. The limit dim(X n ) n→∞ n

stabdim(X ) := lim

is called the stable topological dimension of X . In the case where X is a compact metrizable space with dim(X ) = ∞, we put stabdim(X ) = ∞. We have that dim(X n ) by (6.2.1). In particular, we always have stabdim(X ) = inf n≥1 n stabdim(X ) ≤ dim(X ).

(6.2.3)

Note that this inequality is in fact an equality when X is a polyhedron by Corollary 3.5.11. Remark 6.2.5 If (u n ) is a subadditive sequence of real numbers of arbitrary sign, it un may happen that the sequence ( ) is divergent. For example, the sequence u n = n un 2 = −∞. −n is subadditive but one has limn→∞ n

6.3 Definition of Mean Topological Dimension Let X be a topological space and T : X → X a continuous map. Denote by Id X the identity map on X . The iterates of T are the maps T n : X → X , n ∈ N, inductively defined by T 0 := Id X and T n+1 := T ◦ T n for all n ∈ N. Thus, we have T n = T ◦ T ◦· · · ◦ T n times

for all n ∈ N. Let α be a finite open cover of X . For each integer n ≥ 1, define the finite open cover ω(α, T, n) of X by ω(α, T, n) := α ∨ T −1 (α) ∨ T −2 (α) ∨ · · · ∨ T −n+1 (α) =

n−1 k=0

T −k (α),

6.3 Definition of Mean Topological Dimension

and let

111

D(α, T, n) := D ω(α, T, n) .

Proposition 6.3.1 Let X be a normal space, T : X → X a continuous map, and α a finite open cover of X . Then the sequence (D(α, T, n))n≥1 is subadditive. Proof Let n and m be positive integers. Then, by Proposition 6.1.1, we have that ω(α, T, n + m) =

n+m−1

T −k (α)

k=0

= =

n−1 k=0 n−1

T

−k

(α) ∨

n+m−1

T

−k

(α) ∨ T

T

−k

(α)

k=n −n

m−1

k=0

T

−k

(α)

k=0

= ω(α, T, n) ∨ T −n (ω(α, T, m)). As X is normal, we can apply Proposition 6.1.5. We deduce that D(α, T, n + m) = D ω(α, T, n + m) = D ω(α, T, n) ∨ T −n (ω(α, T, m)) ≤ D ω(α, T, n) + D T −n (ω(α, T, m)) , which implies, by using Proposition 4.4.2, D(α, T, n + m) ≤ D ω(α, T, n) + D ω(α, T, m) = D(α, T, n) + D(α, T, m). Consequently, the sequence (D(α, T, n))n≥1 is subadditive.

Let X be a normal space and T : X → X a continuous map. Let α be a finite open cover of X . By Propositions 6.3.1 and 6.2.3, the limit D(α, T ) := lim

n→∞

D(α, T, n) n

(6.3.1)

exists and is finite. Definition 6.3.2 Let X be a normal space and T : X → X a continuous map. The mean topological dimension of the dynamical system (X, T ) is the quantity mdim(X, T ) defined by

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6 Mean Topological Dimension for Continuous Maps

mdim(X, T ) := sup D(α, T ) ∈ [0, ∞], α

where α runs over all finite open covers of X and D(α, T ) is the non-negative real number defined by formula (6.3.1). Example 6.3.3 Take as T : X → X the identity map. Let α be a finite open cover of X . Then ω(α, T, n) = α ∨ · · · ∨ α for every n. As ω(α, T, n) α and α ω(α, T, n), we deduce from Proposition 1.1.14 that D(α, T, n) = D(α). It follows that D(α, T ) = 0. Thus we have that mdim(X, T ) = 0. Remark 6.3.4 Examples of dynamical systems with positive mean topological dimension will be given in Sect. 7.2.

6.4 General Properties of Mean Topological dimension Let X and X be topological spaces. Let T : X → X and T : X → X be continuous maps. The dynamical systems (X, T ) and (X , T ) are called topologically conjugate if there exists a homeomorphism h : X → X such that T = h ◦ T ◦ h −1 . One then says that the homeomorphism h conjugates T and T . Note that the equality T = h ◦ T ◦ h −1 is equivalent to h ◦ T = T ◦ h, i.e., to the commutativity of the following diagram: T

X −−−−→ ⏐ ⏐ h

X ⏐ ⏐ h

X −−−− → X T

Mean topological dimension is an invariant of topological conjugacy. More precisely, we have the following statement: Proposition 6.4.1 Let X and X be normal spaces. Let T : X → X and T : X → X be continuous maps. Suppose that the dynamical systems (X, T ) and (X , T ) are topologically conjugate. Then one has mdim(X, T ) = mdim(X , T ). Proof Let h : X → X be a homeomorphism that conjugates T and T . Let α be a finite open cover of X and α := h −1 (α ). As h ◦ T = T ◦ h, the homeomorphism h sends ω(α, T, n) to ω(α , T , n) for every n ≥ 1. It follows that D(α, T, n) = D(α , T , n) for all n and hence D(α, T ) = D(α , T ). Since α → α provides a bijective correspondence between the finite open covers of X and those of X , we deduce that mdim(X, T ) = mdim(X , T ). Proposition 6.4.2 Let X be a normal space and T : X → X a continuous map. Then one has

6.4 General Properties of Mean Topological dimension

113

mdim(X, T n ) = n mdim(X, T ) for every integer n ≥ 0. Proof Let α be a finite open cover of X . For every integer k ≥ 1, the cover ω(α, T, kn) = α ∨ T −1 (α) ∨ T −2 (α) ∨ · · · ∨ T −kn+1 (α) is finer than the cover ω(α, T n , k) = α ∨ T −n (α) ∨ T −2n (α) ∨ · · · ∨ T −kn+n (α). Thus, it follows from Proposition 1.1.4 that D(α, T, kn) ≥ D(α, T n , k), which implies n D(α, T ) = lim

k→∞

D(α, T, kn) D(α, T n , k) ≥ lim = D(α, T n ). k→∞ k k

We deduce that n mdim(X, T ) ≥ mdim(X, T n ).

(6.4.1)

On the other hand, by Proposition 6.1.1, we have that ω(α, T, kn) = ω(α, T, n) ∨ T −n (ω(α, T, n)) ∨ · · · ∨ T −(k−1)n (ω(α, T, n)), that is, by setting β := ω(α, T, n), ω(α, T, kn) = ω(β, T n , k). We then get n D(α, T ) = D(β, T n ) ≤ mdim(X, T n ), and hence n mdim(X, T ) ≤ mdim(X, T n ).

(6.4.2)

Inequalities (6.4.1) and (6.4.2) imply that mdim(X, T n ) = n mdim(X, T ).

Corollary 6.4.3 Let X be a normal space and let T : X → X be a homeomorphism. Then one has (6.4.3) mdim(X, T n ) = |n| mdim(X, T )

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6 Mean Topological Dimension for Continuous Maps

for all n ∈ Z. Proof This follows directly from Proposition 6.4.2 if n ≥ 0. Therefore, it suffices to verify Formula (6.4.3) for n = −1. For every finite open cover α of X and every integer k ≥ 1, we have that ω(α, T −1 , k) = T k−1 (ω(α, T, k)). As T k−1 is a homeomorphism, we deduce that D(α, T −1 , k) = D(α, T, k) for all k ≥ 1. This implies D(α, T −1 ) = D(α, T ). It follows that mdim(X, T −1 ) = mdim(X, T ). Proposition 6.4.4 Let X be a normal space with dim(X ) < ∞. Let T : X → X be a continuous map. Then one has mdim(X, T ) = 0. Proof For every finite open cover α of X , we have that D(α, T, n) ≤ dim(X ) by definition of dim(X ). As dim(X ) < ∞, we deduce that D(α, T ) = lim

n→∞

D(α, T, n) = 0. n

Thus, we have that mdim(X, T ) = sup D(α, T ) = 0. α

Let X be a topological space and T : X → X a continuous map. A subset Y ⊂ X is called T -invariant if it satisfies T (Y ) ⊂ Y . If Y ⊂ X is T -invariant, then T induces by restriction a continuous map T |Y : Y → Y given by T |Y (y) = T (y) for all y ∈ Y . Recall from Proposition 1.5.6 that every closed subset of a normal space is itself normal. Proposition 6.4.5 Let X be a normal space and let T : X → X be a continuous map. Let Y ⊂ X be a closed (and hence normal) T -invariant subset of X . Then one has mdim(Y, T |Y ) ≤ mdim(X, T ). Proof Let α = (Ui )i∈I be a finite open cover of Y . For each i ∈ I , we can find an open subset Vi of X such that Ui = Vi ∩ Y . Consider the finite open cover β of X defined by β := (Vi )i∈I ∪ (X \ Y ). Let γ = (W j ) j∈J be a finite open cover of X that is finer than ω(β, T, n). Then γ := (W j ∩ Y ) j∈J is clearly a finite open cover of Y that is finer than ω(α, T |Y , n) and we have ord(γ ) ≤ ord(γ) (cf. the proof of Proposition 1.2.1). It follows that D(α, T |Y , n) ≤ D(β, T, n) for

6.4 General Properties of Mean Topological dimension

115

all n ≥ 1. This implies D(α, T |Y ) ≤ D(β, T ) ≤ mdim(X, T ). We deduce that mdim(Y, T |Y ) = supα D(α, T |Y ) ≤ mdim(X, T ). Remark 6.4.6 In the sequel, when Y is a T -invariant subset, we will sometimes write by abuse T instead of T |Y if there is no risk of confusion. Corollary 6.4.7 Let X be a compact space and Y a normal Hausdorff space. Let T : X → X and S : Y → Y be continuous maps. Suppose that there exists an injective continuous map f : X → Y such that f ◦ T = S ◦ f . Then one has mdim(X, T ) ≤ mdim(Y, S). Proof It follows from our hypotheses that Z := f (X ) is a closed S-invariant subset of Y ant that f induces a topological conjugacy between (X, T ) and (Z , S| Z ). Thus, we have that mdim(X, T ) = mdim(Z , S| Z ) ≤ mdim(Y, S)

by virtue of Propositions 6.4.1 and 6.4.5.

6.5 Metric Approach to Mean Topological Dimension In this section, (X, d) is a compact metric space and T : X → X a continuous map. Let n ≥ 1 be an integer. For all x, y ∈ X , define dn (x, y) by dn (x, y) = dnT (x, y) :=

max d(T k (x), T k (y)).

0≤k≤n−1

Clearly dn is a metric on X . Proposition 6.5.1 The metric dn defines the same topology as d on X . Proof We have that d(x, y) ≤ dn (x, y) for all x, y ∈ X . Therefore, the identity map (X, dn ) → (X, d) is continuous. On the other hand, the continuity of T : (X, d) → (X, d) implies that, if (yi ) is a sequence of points of X such that the sequence d(x, yi ) converges to 0, then the sequence dn (x, yi ) converges also to 0. Consequently, the identity map (X, d) → (X, dn ) is continuous. By Proposition 4.6.2, we have that dimε (X, dn ) < ∞ for all ε > 0 and n ≥ 1. Remark 6.5.2 As dn ≤ dn+1 for all n ≥ 1, we deduce from Corollary 4.6.4 that, if we fix ε > 0, then the sequence (dimε (X, dn ))n≥1 is non-decreasing. Proposition 6.5.3 Let ε > 0. Then the sequence (u n )n≥1 defined by u n := dimε (X, dn ) is subadditive.

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6 Mean Topological Dimension for Continuous Maps

Proof Let n and m be positive integers. Let K be a compact metrizable space such that there exists a continuous map f : X → K that is ε-injective for the metric dn . Let L be a compact metrizable space such that there exists a continuous map g : X → L that is ε-injective for the metric dm . Then the map F : X → K × L defined by F(x) = f (x), g(T n (x)) is clearly continuous and ε-injective for the metric dn+m . We deduce that dimε (X, dn+m ) ≤ dim(K × L) since K × L, being the product of two compact and metrizable spaces, is itself compact and metrizable. As dim(K × L) ≤ dim(K ) + dim(L) by Corollary 4.5.6, this implies dimε (X, dn+m ) ≤ dim(K ) + dim(L) and hence dimε (X, dn+m ) ≤ dimε (X, dn ) + dimε (X, dm ). Consequently, the sequence (u n ) is subadditive.

For every ε > 0, we define the real number mdimε (X, d, T ) ≥ 0 by mdimε (X, d, T ) := lim

n→∞

dimε (X, dn ) ∈ [0, ∞[. n

The above limit exists and is finite by Propositions 6.2.3 and 6.5.3. For a fixed n, the map ε → dimε (X, dn ) is non-increasing. It follows that the map ε → mdimε (X, d, T ) is also non-increasing. We deduce that mdimε (X, d, T ) has a (possibly infinite) limit as ε tends to 0. Theorem 6.5.4 Let (X, d) be a compact metric space and let T : X → X be a continuous map. Then one has mdim(X, T ) = lim mdimε (X, d, T ). ε→0

Proof Consider a finite open cover α of X . Let λ > 0 be a Lebesgue number of α relative to the metric d. We claim that D(α, T, n) ≤ dimλ (X, dn )

(6.5.1)

for every n ≥ 1. Indeed, consider a compact metrizable space K such that there exists a continuous map f : X → K that is λ-injective relatively to the metric dn . Let y ∈ K . As f is λ-injective relatively to dn , the d-diameter of T k ( f −1 (y)) is less than or equal to λ for every integer k such that 0 ≤ k ≤ n − 1. As λ is a Lebesgue number of the cover α, it follows that f −1 (y) is entirely contained in one of the elements of the cover ω(α, T, n) = α ∨ T −1 (α) ∨ · · · ∨ T −n+1 (α). By applying Lemma 4.5.3, we deduce that f is ω(α, T, n)-compatible. Therefore, by using Proposition 4.4.5, we obtain

6.5 Metric Approach to Mean Topological Dimension

117

D(α, T, n) = D(ω(α, T, n)) ≤ dim(K ), which yields (6.5.1). Inequality (6.5.1) implies D(α, T ) = lim

n→∞

D(α, T, n) dimλ (X, dn ) ≤ lim = mdimλ (X, d, T ). n→∞ n n

As the map ε → mdimε (X, d, T ) is non-increasing, we deduce that D(α, T ) ≤ lim mdimε (X, d, T ), ε→0

and hence, by taking the upper bound over α, mdim(X, T ) ≤ lim mdimε (X, d, T ). ε→0

To complete the proof, it suffices to establish that lim mdimε (X, d, T ) ≤ mdim(X, T ).

ε→0

(6.5.2)

Let ε > 0. Consider the open cover of X by its open d-balls of radius ε/2. By compactness of X , it admits a finite subcover α. Let n be a positive integer. Consider the finite open cover ω(α, T, n). By Proposition 4.4.6, we can find a polyhedron P such that dim(P) = D(α, T, n) and a continuous ω(α, T, n)-compatible map f : X → P. Let y ∈ P. As f is ω(α, T, n)-compatible, the set f −1 (y) is contained in one of the open sets of the cover ω(α, T, n). Consequently, for each integer k such that 0 ≤ k ≤ n − 1, the set T k ( f −1 (y)) is contained in one of the open balls of radius ε/2. Thus, the map f is ε-injective for the metric dn . As P is compact and metrizable, we deduce that dimε (X, dn ) ≤ dim(P) = D(α, T, n). We then get mdimε (X, d, T ) = lim

n→∞

dimε (X, dn ) D(α, T, n) ≤ lim = D(α, T ). n→∞ n n

Since D(α, T ) ≤ mdim(X, T ), we obtain mdimε (X, d, T ) ≤ mdim(X, T ), which yields (6.5.2) after letting ε tend to 0.

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6 Mean Topological Dimension for Continuous Maps

Notes Subadditivity plays an important role in many branches of pure and applied mathematics. Fekete’s lemma (Proposition 6.2.3), which is named after Fekete (cf. [35, Satz 2] and also [90, p. 198]) has been generalized in various directions. For example, it is known [63, Theorem 16.2.9] that if f is a measurable subadditive realvalued map on Rn then, for every x ∈ Rn , the function g(t) := f (t x)/t admits inf t>0 g(t) ∈ R ∪ {−∞} as a limit as t tends to infinity. There exist compact metrizable spaces X for which Inequality (6.2.3) is strict. Actually, Boltyanskiˇı [15, 16] gave examples of compact metrizable spaces X satisfying dim(X ) = 2 and dim(X × X ) = 3. For such a space X , we have that stabdim(X ) ≤ 3/2 < dim(X ). Since the inequality dim(X × Y ) ≤ dim(X ) + dim(Y ) remains valid whenever X and Y are compact Hausdorff or metrizable (see the Notes on dim(X n ) exists and thus the definition of stabdim(X ) Chap. 4), the limit limn→∞ n may be extended to the case when X is compact Hausdorff or metrizable. Mean topological dimension was introduced by Gromov [44] for studying dynamical properties of certain spaces of holomorphic maps and complex varieties. It was used by Lindenstrauss and Weiss [74] to answer in the negative a question that had been raised by Auslander (see Chap. 8). The paper by Lindenstrauss and Weiss contains many other important results about mean topological dimension. It is shown in particular in [74] that if T is a homeomorphism of a compact metrizable space X such that (X, T ) is uniquely ergodic or has finite topological entropy, then mdim(X, T ) = 0 (cf. Exercise 6.11 for the definition of topological entropy).

Exercises 6.1 Let a be a real number such that 0 ≤ a ≤ 1. Show that the sequence (n a )n≥1 is subadditive. 6.2 Let C be a positive real number. Show that the sequence (log(C + n))n≥1 is subadditive. 6.3 Let T : X → X be a map from a set X into itself and let F be a finite subset of X . For each integer n ≥ 1, let u n denote the cardinality of the set F ∪ T (F) ∪ T 2 (F) ∪ · · · ∪ T n−1 (F) =

n−1

T k (F).

k=0

(a) Show that the sequence (u n )n≥1 is subadditive. (b) Show that the sequence (vn )n≥1 , defined by vn := u n+1 − u n for all n ≥ 1, is non-increasing.

Exercises

119

(c) Show that there exist integers α ≥ 0 and n 0 ≥ 1 such that vn = α for all n ≥ n0. un = α. (d) Show that limn→∞ n 6.4 Let S be a semigroup, i.e., a set equipped with an associative binary operation. Let A be a non-empty subset of S. For n ≥ 1, denote by γn the number of elements s ∈ S that can be written in the form s = a1 a2 . . . ak with k ≤ n and ai ∈ A for all 1 ≤ i ≤ k. Show that the sequence (u n )n≥1 defined by u n := log γn is subadditive. 6.5 Let (u n )n≥1 be a subadditive sequence of real numbers. Show that if the seun un quence ( ) is not convergent then one has limn→∞ = −∞. n n 6.6 Let (u n )n≥1 be a sequence of real numbers. Suppose that there exists a real number C such that u n+m ≤ u n + u m + C for all n, m ≥ 1. Show that the sequence (u n + C) is subadditive. Deduce that un the sequence ( ) either is convergent or has −∞ as a limit. n 6.7 (Translation number). Let f : R → R be a homeomorphism such that f (x + k) = f (x) + k for all x ∈ R and k ∈ Z. Let x ∈ R. Show that the limit τ ( f ) := lim

n→∞

f n (x) n

exists and is finite and that this limit does not depend on the choice of the point x ∈ R. Hint: first observe that f is increasing and then prove that the sequence ( f n (0) + 1)n≥1 is subadditive. 6.8 Let G be a group. A map q : G → R is called a quasi-homomorphism if the map (g1 , g2 ) → q(g1 g2 ) − q(g1 ) − q(g2 ) is bounded on G × G. Let q : G → R be a quasi-homomorphism. q(g n ) ) is convergent. n (b) Consider the map q∞ : G → R defined by (a) Let g ∈ G. Show that the sequence (

q(g n ) . n→∞ n

q∞ (g) := lim

Show that q∞ is a quasi-homomorphism. (c) Show that q∞ (g n ) = nq∞ (g) and q∞ (h −1 gh) = q∞ (g) for all n ∈ Z and g, h ∈ G. 6.9 A map f : [0, ∞) → R is called subadditive if it satisfies f (x + y) ≤ f (x) + f (y) for all x, y ∈ [0, ∞). (a) Let f : [0, ∞) → R be a continuous subadditive map. By adapting the proof of Proposition 6.2.3, show that the map g : (0, ∞) → R defined by

120

6 Mean Topological Dimension for Continuous Maps

f (x) has a limit λ ∈ R ∪ {−∞} as x → ∞ and that one has x λ = inf x>0 g(x). (b) Recall that a map f : [0, ∞) → R is called concave if it satisfies f ((1 − t)x + t y) ≥ (1 − t) f (x) + t f (y) for all x, y ∈ [0, ∞) and t ∈ [0, 1]. Show that a concave map f : [0, ∞) → R is subadditive if and only if f (0) ≥ 0. (c) Show that there exists a non-linear map f : R → R such that f (x + y) = f (x) + f (y) for all x, y ∈ R. Hint: use the fact that R, viewed as a vector space over the field Q, admits a base. f (x) has no limit as (d) Show that if f is as in the previous question then x x → ∞. (This shows that the hypothesis that f is continuous cannot be removed in the first question.) g(x) :=

6.10 Let X be a normal space and T : X → X a continuous map. Let α and β be finite open covers of X . Show that one has D(α∨β, T ) ≤ D(α, T )+ D(β, T ). 6.11 Let X be a non-empty topological space and T : X → X a continuous map. Given a finite open cover α = (Ui )i∈I of X , denote by N (α) the smallest integer k ≥ 1 such that there exists a subset I0 ⊂ I with cardinality k satisfying i∈I0 Ui = X . (a) Let α and β be finite open covers of X . Show that N (α ∨β) ≤ N (α)N (β). (b) Let α be a finite open cover of X and f : X → X a continuous map. Show that N ( f −1 (α)) ≤ N (α). (c) Let α be a finite open cover of X . Given an integer n ≥ 1, let ω(α, T, n) :=

n−1

T −k (α).

k=0

Show that the limit Htop (X, T, α) := lim

n→∞

log N (ω(α, T, n)) n

exists and is finite. Hint: observe that the sequence (log N (ω(α, T, n)))n≥1 is subadditive. The quantity h top (X, T ) := sup Htop (X, T, α), α

where α runs over all finite open covers of X , is called the topological entropy of the dynamical system (X, T ). (d) Let Y be a topological space and S : Y → Y a continuous map. Suppose that there exists a surjective continuous map f : Y → X such that f ◦ S = T ◦ f . Show that one has h top (X, T ) ≤ h top (Y, S).

Exercises

121

6.12 Let X be a normal space and T : X → X a constant map. Show that mdim(X, T ) = 0. 6.13 Let (X, d) be a compact metric space and T : X → X a map satisfying d(T (x), T (y)) ≤ d(x, y) for all x, y ∈ X . Show that mdim(X, T ) = 0. 6.14 Let X 1 and X 2 be compact metrizable spaces. Let T1 : X 1 → X 1 and T2 : X 2 → X 2 be continuous maps. Consider the product map T1 × T2 : X 1 × X 2 → X 1 × X 2 defined by T1 × T2 (x1 , x2 ) := (T1 (x1 ), T2 (x2 )). Show that mdim(X 1 × X 2 , T1 × T2 ) ≤ mdim(X 1 , T1 ) + mdim(X 2 , T2 ). Hint: use Theorem 6.5.4 and the result of Exercise 4.7. 6.15 Let X and Y be compact metrizable spaces. Let T : X → X be a continuous map. Show that mdim(X × Y, T × IdY ) = mdim(X, T ). 6.16 Let (X, d) be a compact metric space and let T : X → X be a continuous map. For n ≥ 1, let dn be the metric on X defined by dn (x, y) := max0≤k≤n−1 d(T k (x), T k (y)). (a) (cf. Exercises 4.11 and 3.11). Let ε > 0. Show that the sequence (Widimε (X, dn ))n≥1 is subadditive. (b) Show that the limit mWidimε (X, d, T ) := lim

n→∞

Widimε (X, dn ) n

exists and is finite. (c) Show that mdim(X, T ) = lim mWidimε (X, d, T ). ε→0

Chapter 7

Shifts and Subshifts over Z

In this chapter, we introduce the shift map σ : K Z → K Z on the space of bi-infinite sequences of points in a topological space K . We prove that mdim(K Z , σ) ≤ dim(K ) whenever K is compact and metrizable (Theorem 7.1.3) and that equality holds if K is a polyhedron (Corollary 7.2.3). In the last section, we prove the existence, for any real number 0 ≤ λ ≤ 1, of a closed shift-invariant subset X ⊂ [0, 1]Z with mean topological dimension mdim(X, σ) = λ (Theorem 7.6.2). As an application, we show that mean topological dimension can take any value in [0, ∞] (Corollary 7.6.5).

7.1 Shifts Let K be a topological space. Consider theset K Z formed by all bi-infinite sequences (xi )i∈Z of points of K . We equip K Z = i∈Z K with its product topology (i.e., the coarsest topology for which the projection map K Z → K given by x → xi is continuous for every i ∈ Z). The map σ = σ K : K Z → K Z that sends the sequence x = (xi ) ∈ K Z to the sequence σ(x) := (yi ), where yi = xi+1 for all i ∈ Z is called the shift map on K Z . Clearly σ is a homeomorphism of K Z with inverse σ −1 : K Z → K Z given by σ −1 (x) = (z i ), where z i = xi−1 for all i ∈ Z. The dynamical system (K Z , σ) is called the full shift, or simply the shift, over Z with symbol space K . Our goal is to evaluate the mean topological dimension mdim(K Z , σ) of this dynamical system. Recall that mdim(X, T ) has been only defined in the case when the ambient space X is normal. The following example shows that it may happen that K is normal and K Z is not. Example 7.1.1 The Sorgenfrey line S is normal by Corollary 5.5.3 but its square S × S is not by Proposition 5.5.9. As S × S clearly embeds as a closed subset of S Z , it then follows from Proposition 1.5.6 that S Z is not normal. Note however that if K is metrizable then K Z is also metrizable and hence normal by Proposition 1.5.3. Indeed, the product of a countable family of metrizable spaces © Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_7

123

7 Shifts and Subshifts over Z

124

is metrizable. To see this in our particular setting, suppose that K is a metrizable space and let d be a metric on K that is compatible with the topology. As the metric min(1, d K ) is also compatible with the topology, we may assume that d K is bounded. Then one easily verifies that the metric d on K Z , given by d(x, y) =

∞ d K (xi , yi ) , 2|i|

(7.1.1)

i=−∞

for all x = (xi ), y = (yi ) ∈ K Z , is compatible with the topology on K Z . Another class of topological spaces for which K Z is normal whenever K is in the class is that formed by compact Hausdorff spaces. Indeed, any product of compact Hausdorff spaces is itself a compact Hausdorff space and hence normal by Proposition 1.5.4. Thus, the mean topological dimension mdim(K Z , σ) is well defined if K is a metrizable space or a compact Hausdorff space. Example 7.1.2 Let K be a compact Hausdorff space or a separable metrizable space such that dim(K ) = 0 (e.g., a non-empty finite set with its discrete topology or a space homeomorphic to the Cantor ternary set). As dim(K Z ) = 0 by Corollaries 2.4.24 and 2.4.28, we deduce from Proposition 6.4.4 that mdim(K Z , σ) = 0. Recall that the stable topological dimension of a non-empty compact metrizable space K is the quantity dim(K n ) dim(K n ) = inf ∈ [0, ∞]. n→∞ n≥1 n n

stabdim(K ) = lim (see Example 6.2.4).

Theorem 7.1.3 Let K be a non-empty compact metrizable space. Then one has mdim(K Z , σ) ≤ stabdim(K ) ≤ dim(K ).

(7.1.2)

Proof Let d K be a metric on K compatible with the topology and consider the metric d on K Z defined by Formula (7.1.1) (observe that d is bounded by compactness of K ). Fix some ε > 0. Let δ := diam(K ) and choose an integer r ≥ 0 such that δ < ε. 2r −1 Let n ≥ 1 be an integer and dn the metric on K Z defined by dn (x, y) :=

max d(σ k (x), σ k (y)).

0≤k≤n−1

7.1 Shifts

125

Consider the continuous map f : K Z → K 2r +n given by f (x) := (x−r , x−r +1 , . . . , xr +n−1 ) for all x = (xi ) ∈ K Z . Let x = (xi ) and y = (yi ) be two sequences in K Z such that f (x) = f (y). This means that xi+k = yi+k for all i ∈ {−r, . . . , r } and k ∈ {0, . . . , n − 1}. In other words, for any k ∈ {0, . . . , n − 1}, the terms with index i of the sequences σ k (x) and σ k (y) coincide if i ∈ {−r, . . . , r }. We deduce that d(σ k (x), σ k (y)) ≤ 2

∞ δ δ = r −1 < ε, i 2 2

i=r +1

for every k ∈ {0, . . . , n − 1}. This implies dn (x, y) < ε. Consequently, f is εinjective with respect to the metric dn . As K 2r +n is a compact metrizable space, we deduce that (7.1.3) dimε (K Z , dn ) ≤ dim(K 2r +n ). This yields dimε (K Z , dn ) n→∞ n dim(K 2r +n ) ≤ lim n→∞ n dim(K 2r +n ) = lim n→∞ 2r + n = stabdim(K ),

mdimε (K Z , d, σ) = lim

and hence, by using Theorem 6.5.4, mdim(K Z , σ) = lim mdimε (K Z , d, σ) ≤ stabdim(K ) ≤ dim(K ). ε→0

7.2 Shifts on Polyhedra The following result provides examples of dynamical systems with finite positive mean topological dimension. Theorem 7.2.1 Let N ∈ N and let K := [0, 1] N be the N -dimensional cube. Then one has mdim(K Z , σ) = N .

7 Shifts and Subshifts over Z

126

Proof We have that mdim(K Z , σ) ≤ dim(K ) by Theorem 7.1.3. As dim(K ) = N by Theorem 3.5.4, it only remains to establish the inequality mdim(K Z , σ) ≥ N . Let d K be the metric on K associated with the sup-norm · ∞ on R N and d the metric on K Z defined by Formula (7.1.1). Observe that (7.2.1) d K (x0 , y0 ) ≤ d(x, y) for all x = (xi ), y = (yi ) ∈ K Z . Let n ≥ 1 be an integer. Consider the metric dn on K Z defined by dn (x, y) :=

max d σ k (x), σ k (y) .

0≤k≤n−1

Inequality (7.2.1) yields max d K (xk , yk ) ≤ dn (x, y)

0≤k≤n−1

(7.2.2)

for all x, y ∈ K Z . Consider now the topological embedding ϕ : K n −→ K Z that sends each u = (u 1 , . . . , u n ) ∈ K n to the sequence (xi ) ∈ K Z defined by u i+1 xi = 0

if 0 ≤ i ≤ n − 1, otherwise.

Denoting by ρ the metric induced by the sup-norm · ∞ on K n = [0, 1]n N ⊂ Rn N , Inequality (7.2.2) implies ρ(u, v) ≤ dn ϕ(u), ϕ(v) for all u, v ∈ K n . By applying Proposition 4.6.3, we deduce that dimε (K n , ρ) ≤ dimε (K Z , dn ) for all ε > 0. As dimε (K n , ρ) = n N for all ε ≤ 1 by Proposition 4.6.5, we obtain n N ≤ dimε (K Z , dn ) for all ε ≤ 1. Consequently, we get dimε (K Z , dn ) ≥N n→∞ n

mdimε (K Z , d, σ) = lim

7.2 Shifts on Polyhedra

127

for all ε ≤ 1. By applying Theorem 6.5.4, we deduce that mdim(K Z , σ) = lim mdimε (K Z , d, σ) ≥ N . ε→0

This shows (7.2).

Corollary 7.2.2 Let N ∈ N and let K be a metrizable or compact Hausdorff space such that there exists a subset A ⊂ K that is homeomorphic to the N -cube [0, 1] N . Then one has mdim(K Z , σ) ≥ N . Proof The subset AZ ⊂ K Z is closed and σ-invariant. It follows that mdim(K Z , σ) ≥ mdim(AZ , σ) = N by Proposition 6.4.5. Corollary 7.2.3 Let P be a polyhedron. Then one has mdim(P Z , σ) = dim(P). Proof Let N := dim(P). Since every polyhedron is compact and metrizable, we have that mdim(P Z , σ) ≤ N by Theorem 7.1.3. As P is a polyhedron, we can find n ∈ N and a simplicial complex C of Rn such that P is homeomorphic to |C|. We know that the combinatorial dimension of C is equal to N by Corollary 3.5.5. Therefore, the complex C contains an N -simplex. This shows that we can find a subset A ⊂ P that is homeomorphic to [0, 1] N . Thus, mdim(P Z , σ) ≥ N by Corollary 7.2.2. Corollary 7.2.4 Let K = [0, 1]N be the Hilbert cube. Then one has mdim(K Z , σ) = ∞. Proof The subset A ⊂ K consisting of all (u n )n∈N ∈ K such that u n = 0 for all n ≥ N is homeomorphic to [0, 1] N . Thus, we deduce from Corollary 7.2.2 that mdim(K Z , σ) ≥ N for all N ∈ N.

7.3 Mean Projective Dimension of Subshifts Let K be a topological space. Consider the shift map σ : K Z → K Z . A subset X ⊂ K Z is called a subshift if X is a closed subset of K Z and σ(X ) = X . Example 7.3.1 Let A := {σ n (x) | n ∈ Z} denote the σ-orbit of a point x ∈ K Z . Then its closure X = A is a subshift of K Z . This subshift is called the orbit closure of x. Example 7.3.2 The intersection of any family of subshifts of K Z is itself a subshift of K Z . Example 7.3.3 Any finite union of subshifts of K Z is itself a subshift of K Z .

7 Shifts and Subshifts over Z

128

For each integer n ≥ 1, we denote by wn the projection map from K Z on K n defined by (7.3.1) wn (x) := (x0 , x1 , . . . , xn−1 ) for all x = (xi ) ∈ K Z . Proposition 7.3.4 Let K be a compact metrizable space with topological dimension dim(K ) < ∞. Let X ⊂ K Z be a non-empty subshift. Then the sequence (u n )n≥1 defined by u n := dim(wn (X )) is subadditive. Proof Let n and m be positive integers. As X is σ-invariant, we have that wn+m (X ) ⊂ wn (X ) × wm (X ) ⊂ K n × K m = K n+m . By applying Proposition 1.2.1 and Corollary 4.5.6, we deduce that dim(wn+m (X )) ≤ dim(wn (X )) + dim(wm (X )), which shows that the sequence (u n ) is subadditive.

Suppose that K is a compact metrizable space with topological dimension dim(K ) < ∞ and that X ⊂ K Z is a non-empty subshift. As wn (X ) ⊂ K n , we have that dim(wn (X )) ≤ n dim(K ) < ∞ for all n ≥ 1. Thus, it follows from Propositions 7.3.4 and 6.2.3 that the limit prodim(X ) := lim

n→∞

dim(wn (X )) n

(7.3.2)

exists and is finite. One says that prodim(X ) is the mean projective dimension of the subshift X . Note that dim(K n ) = stabdim(K ) ≤ dim(K ), n→∞ n

prodim(X ) ≤ prodim(K Z ) = lim

since wn (X ) ⊂ K n for all n. It turns out that the mean projective dimension of a subshift X ⊂ K Z yields an upper bound for the mean topological dimension of the dynamical system (X, σ). More precisely, we have the following result, which extends Theorem 7.1.3. Theorem 7.3.5 Let K be a compact metrizable space with dim(K ) < ∞ and X ⊂ K Z a non-empty subshift. Then one has mdim(X, σ) ≤ prodim(X ). Proof Using the notation introduced in the proof of Theorem 7.1.3, the map f : K Z → K 2r +n induces by restriction a continuous map g : X → w2r +n (X ).

7.3 Mean Projective Dimension of Subshifts

129

As f is ε-injective with respect to the metric dn , the same is true for g. Therefore we have that dimε (X, dn ) ≤ dim(w2r +n (X )). By using the result of Proposition 7.3.4, we deduce that dimε (X, dn ) ≤ dim(w2r (X )) + dim(wn (X )), and hence dimε (X, dn ) n dim(wn (X )) = prodim(X ). ≤ lim n→∞ n

mdimε (X, d, σ) = lim

n→∞

Finally, we get mdim(X, σ) = lim mdimε (X, d, σ) ≤ prodim(X ). ε→0

7.4 Subshifts of Finite Type Let K be a topological space. Let q ≥ 0 be an integer and L a closed subset of K q . Consider the subset X ⊂ K Z consisting of all sequences (xi ) ∈ K Z such that (xk , xk+1 , . . . , xk+q−1 ) ∈ L for all k ∈ Z. Note that X = {x ∈ K Z | wq (σ k (x)) ∈ L for all k ∈ Z} σ −k wq−1 (L), = k∈Z

where wq : K Z → K q is the projection map defined by (7.3.1). Clearly X is a subshift of K Z . One says that X is the subshift of finite type associated with the pair (q, L). One also says that (q, L) is a defining law for X . Proposition 7.4.1 Let K be a compact metrizable space with dim(K ) < ∞. Let q ≥ 0 be an integer and L a non-empty closed subset of K q . Let X ⊂ K Z denote the subshift of finite type with defining law (q, L). Then one has prodim(X ) ≤

stabdim(L) dim(L) ≤ . q q

(7.4.1)

7 Shifts and Subshifts over Z

130

Proof As wqn (X ) ⊂ L n for every n ≥ 1, we have that dim(wqn (X )) 1 stabdim(L) dim(L n ) ≤ lim = . n→∞ qn q n→∞ n q

prodim(X ) = lim

This yields (7.4.1) since stabdim(Y ) ≤ dim(Y ) for any compact metrizable space Y . Example 7.4.2 Let K be a non-empty polyhedron and q ≥ 1 an integer. Take L = K q . The subshift of finite type X ⊂ K Z with defining law (q, L) it the full shift K Z . By applying Corollaries 7.2.3 and 3.5.10, we see that we have the equalities dim(L) mdim(X, σ) = prodim(X ) = = dim(K ). q Example 7.4.3 Take K = [0, 1], q = 2 and L = [0, 1] × {0} ⊂ [0, 1]2 = K q . The subshift of finite type X ⊂ K Z with defining law (q, L) is reduced to the identically-zero sequence. In this case, we have that mdim(X, σ) = prodim(X ) = 0 <

dim(L) 1 stabdim(L) = = . q q 2

7.5 Subshifts of Block-Type Let K be a topological space. Let q ≥ 1 be an integer and B a closed subset of K q . Define the subset X 0 ⊂ K Z by X 0 := {x = (xi )i∈Z ∈ K Z | (xkq , xkq+1 , . . . , xkq+q−1 ) ∈ B for all k ∈ Z}. (7.5.1) Clearly X 0 is σ q -invariant. Observe that X 0 = {x ∈ K Z | wq (σ kq (x)) ∈ B for all k ∈ Z} σ −kq wq−1 (B). = k∈Z

This shows that X 0 is the intersection of closed subsets of K Z and hence closed in K Z. Proposition 7.5.1 The dynamical system (X 0 , σ q ) is topologically conjugate to the shift on B Z . Proof Consider the map ϕ : X 0 → B Z that sends each x = (xi )i∈Z ∈ X 0 to the sequence y = (yk )k∈Z ∈ B Z defined by yk := (xkq , xkq+1 , . . . , xkq+q−1 )

7.5 Subshifts of Block-Type

131

for all k ∈ Z. Clearly ϕ is a homeomorphism from X 0 onto B Z and satisfies ϕ ◦ σ q = σ B ◦ ϕ. Thus, ϕ conjugates the dynamical systems (X 0 , σ q ) and (B Z , σ B ). Consider now the subset X ⊂ K Z defined by X := {x ∈ K Z | there exists k ∈ Z such that σ k (x) ∈ X 0 }. In other words, the set X consists of all sequences x ∈ K Z that can be obtained by concatenating elements of B. It is clear from this characterization of the elements of X that X is σ-invariant. On the other hand, since σ −m (X 0 ), X= 0≤m≤q−1

the set X is a finite union of closed subsets of K Z and hence closed in K Z . Consequently, X is a subshift of K Z . One says that X is the subshift of block-type of K Z associated with the pair (q, B). Proposition 7.5.2 Let K be a compact metrizable space with dim(K ) < ∞, q ≥ 1 an integer, and B a non-empty closed subset of K q . Let X ⊂ K Z denote the subshift of block-type associated with (q, B). Then one has dim(B) stabdim(B) ≤ . q q

mdim(X, σ) ≤ prodim(X ) =

(7.5.2)

Proof For every n ≥ 1, we have that B n ⊂ wqn (X ) ⊂

K m × B n−1 × K q−m .

0≤m≤q−1

By applying Corollaries 1.2.6 and 4.5.6, we deduce that dim(B n ) ≤ dim(wqn (X )) ≤ dim(B n−1 ) + q dim(K ). After dividing by qn and letting n tend to infinity, we conclude that prodim(X ) =

stabdim(B) . q

This gives us (7.5.2) since mdim(X, σ) ≤ prodim(X ) by Theorem 7.3.5.

Theorem 7.5.3 Let K be a compact metrizable space with dim(K ) < ∞, q ≥ 1 an integer, and B ⊂ K q a polyhedron. Let X ⊂ K Z denote the subshift of block-type associated with (q, B). Then one has mdim(X, σ) =

dim(B) . q

(7.5.3)

7 Shifts and Subshifts over Z

132

Proof The inequality mdim(X, σ) ≤

dim(B) . q

follows from Proposition 7.5.2. On the other hand, consider the closed σ q -invariant subset X 0 ⊂ X defined by (7.5.1). As the dynamical system (X 0 , σ q ) is topologically conjugate to the shift on B Z by Proposition 7.5.1, we have that mdim(X, σ q ) q mdim(X 0 , σ q ) ≥ q

mdim(X, σ) =

(by Proposition 6.4.2) (by Proposition 6.4.5)

mdim(B Z , σ B ) (by Proposition 6.4.1) q dim(B) (by Corollary 7.2.3). = q =

This shows (7.5.3).

7.6 Construction of Subshifts with Prescribed Mean Dimension In this section, we shall prove in particular that the mean topological dimension of a dynamical system can take any value in [0, ∞] (see Corollary 7.6.5). Lemma 7.6.1 Let K := [0, 1] and m ≥ 1 an integer. Let I ⊂ {1, . . . , m} a subset with cardinality r . Let B be the subset of K m defined by / I }. B := {(u 1 , . . . , u m ) ∈ K m | u i = 0 for all i ∈ Let X ⊂ K Z denote the subshift of block-type associated with (m, B). Then one has mdim(X, σ) =

r . m

Proof Since the space B is homeomorphic to [0, 1]r , we deduce from Theorem 7.5.3 that r dim([0, 1]r ) = . mdim(X, σ) = m m It follows from Theorem 7.2.1 and Proposition 6.4.5 that every subshift X ⊂ [0, 1]Z satisfies 0 ≤ mdim(X, σ) ≤ 1. Conversely, we have the following result.

7.6 Construction of Subshifts with Prescribed Mean Dimension

133

Theorem 7.6.2 Let λ be a real number such that 0 ≤ λ ≤ 1. Then there exists a subshift X ⊂ [0, 1]Z such that mdim(X, σ) = λ. Proof Let us choose some integer q ≥ 2 that will serve as a numeration base (for example q = 10 if you are used to count on your fingers). For each integer n ≥ 0, let E n := {0, 1, . . . , q n − 1} denote the set consisting of all non-negative integers that are less than q n . Let an ∈ N denote the integral part of q n λ and let bn := an + 1. Then the sequences (u n ) and (vn ) defined by an bn u n := n and vn := n q q satisfy u n ≤ λ < vn and vn − u n <

1 . qn

It follows that lim u n = lim vn = λ.

n→∞

n→∞

Moreover, the sequence (u n ) (resp. (vn )) is non-decreasing (resp. non-increasing). Let πn : E n+1 → E n denote the map that sends each k ∈ E n+1 to the remainder of the Euclidean division of k by q n . As qan ≤ an+1 ≤ bn+1 ≤ qbn for all n, we can construct by induction on n two sequences (In )n∈N and (Jn )n∈N of subsets of N satisfying the following conditions: (C1) In ⊂ Jn ⊂ E n , (C2) #In = an and # Jn = bn , (C3) πn−1 (In ) ⊂ In+1 and Jn+1 ⊂ πn−1 (Jn ) for all n ≥ 0. Indeed, we can start by taking I0 = ∅ and J0 = {0}. Then, assuming that the subsets In and Jn have already been constructed and that they satisfy (C1) and (C2), we remark that πn−1 (In ) ⊂ πn−1 (Jn ), by (C1). On the other hand, we have that #πn−1 (In ) = qan and #πn−1 (Jn ) = qbn . Thus, we can find sets In+1 and Jn+1 satisfying πn−1 (In ) ⊂ In+1 ⊂ Jn+1 ⊂ πn−1 (Jn ), #In+1 = an+1 and # Jn+1 = bn+1 . n Let K := [0, 1]. Consider the subsets An and Bn of K q defined by n

/ In }, and An := {(u 0 , . . . , u q n −1 ) ∈ K q | u i = 0 if i ∈ n

Bn := {(u 0 , . . . , u q n −1 ) ∈ K q | u i = 0 if i ∈ / Jn }.

7 Shifts and Subshifts over Z

134

Let Yn ⊂ K Z and Z n ⊂ K Z denote the subshifts of block-type associated with (q n , An ) and (q n , Bn ) respectively. Conditions (C1) and (C3) imply that Yn ⊂ Yn+1 ⊂ Z n+1 ⊂ Z n for all n ≥ 0. Consider now the subshift X ⊂ K Z defined by X :=

Zn .

n≥0

It follows from Lemma 7.6.1 and Condition (C2) that an = u n , and qn bn mdim(Z n , σ) = n = vn . q mdim(Yn , σ) =

As X ⊂ Z n , Proposition 6.4.5 gives us mdim(X, σ) ≤ mdim(Z n , σ) = vn for all n ≥ 0. By letting n tend to infinity, we obtain mdim(X, σ) ≤ λ.

(7.6.1)

On the other hand, for every N ≥ n, we have that Yn ⊂ Y N ⊂ Z N . We deduce that Yn ⊂

Z N = X.

N ≥n

By applying again Proposition 6.4.5, this gives us u n = mdim(Yn , σ) ≤ mdim(X, σ) for all n ≥ 0. By letting n tend to infinity, we finally get λ ≤ mdim(X, σ). Inequalities (7.6.1) and (7.6.2) imply that mdim(X, σ) = λ.

(7.6.2)

Corollary 7.6.3 Let n be a positive integer and let K := [0, 1]n be the n-dimensional cube. Let ρ be a real number such that 0 ≤ ρ ≤ n. Then there exists a subshift Y ⊂ K Z such that mdim(Y, σ K ) = ρ.

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135

Proof By Theorem 7.6.2, we can find a subshift X ⊂ [0, 1]Z such that mdim(X, σ) = ρ/n. Now consider the map h : [0, 1]Z → K Z that sends each x = (xi )i∈Z ∈ [0, 1]Z to the sequence y = (yi )i∈Z ∈ K Z defined by yi := (xni , xni+1 , . . . , xni+n−1 ) for all i ∈ Z. Clearly h is a homeomorphism conjugating σ n and σ K . It follows that Y := h(X ) is a subshift of K Z and that the dynamical systems (X, σ n ) and (Y, σ K ) are topologically conjugate. As mdim(Y, σ K ) = mdim(X, σ n ) = n mdim(X, σ)

(by Proposition 6.4.1) (by Proposition 6.4.2)

= ρ, the subshift Y ⊂ K Z has the required properties.

Corollary 7.6.4 Let P be a polyhedron and ρ a real number such that 0 ≤ ρ ≤ dim(P). Then there exists a subshift Z ⊂ P Z such that mdim(Z , σ) = ρ. Proof If n := dim(P), then P contains a subset A homeomorphic to [0, 1]n . By Corollary 7.6.3, we can find a subshift Z ⊂ AZ ⊂ P Z such that mdim (Z , σ) = ρ. Corollary 7.6.5 For every δ ∈ [0, ∞], there exists a dynamical system (X, T ), where X is a compact metrizable space and T : X → X is a homeomorphism, such that mdim(X, T ) = δ. Proof This immediately follows from Corollaries 7.6.3 and 7.2.4.

Notes The branch of the theory of dynamical systems devoted to the investigation of the dynamical properties of shift maps σ : K Z → K Z is known as symbolic dynamics. The most studied case is when the symbol space K is a finite or countably infinite discrete space (see for example [59, 70]). Theorem 7.2.1 is in [44, 74]. Theorem 7.3.5 is a particular case of Proposition 1.9. A in [44]. As mentioned in the Notes on Chap. 6, Boltyanskiˇı [15, 16] gave examples of compact metrizable spaces K satisfying stabdim(K ) < dim(K ). For such spaces K , we have that mdim(K Z , σ) < dim(K ) by Theorem 7.1.3. It would be interesting to find an example of a compact metrizable space K for which mdim(K Z , σ) < stabdim(K ).

7 Shifts and Subshifts over Z

136

Exercises 7.1 Let K be a metrizable space and let d K be a bounded metric on K that is compatible with its topology. Show that if the set K Z is equipped with the metric d defined by Formula (7.1.1), then the shift map σ : K Z → K Z is 2-Lipschitz, i.e., it satisfies d(σ(x), σ(y)) ≤ 2d(x, y) for all x, y ∈ K Z . 7.2 Let X be a topological space and T : X → X a continuous map. The dynamical system (X, T ) is called topologically mixing if, given any two non-empty open subsets U and V of X , there are only finitely many n ∈ Z such that T n (U ) ∩ V = ∅. Let K be a topological space and let σ : K Z → K Z denote the shift map on K Z . Show that the dynamical system (K Z , σ) is topologically mixing. 7.3 Let K and L be topological spaces. Show that the dynamical systems ((K × L)Z , σ K ×L ) and (K Z × L Z , σ K × σ L ) are topologically conjugate. 7.4 Let K := {0, 1} and σ : K Z → K Z the shift map. Let X denote the subset of K Z consisting of all x = (xi )i∈Z ∈ K Z such that there is at most one integer i ∈ Z with xi = 1. (a) Show that X is a subshift of K Z . (b) Show that the subshift X is not of finite type. (c) Show that the dynamical system (X, σ) is not topologically mixing. 7.5 Let K := {0, 1} and σ : K Z → K Z the shift map. Let X denote the subset of K Z consisting of all sequences with no consecutive 1s. Let Y denote the subset of K Z consisting of all sequences such that between any two 1s there are always an even number of 0s. (a) Show that X and Y are subshifts of K Z . The subshifts X is called the golden mean subshift and the subshift Y is called the even subshift. (b) Show that the dynamical systems (X, σ) and (Y, σ) are not topologically conjugate. Hint: count fixed points. (c) Show that the subshift X is of finite type but Y is not. (d) Show that the dynamical systems (X, σ) and (Y, σ) are both topologically mixing. (e) Show that the map f : X → Y that sends each x = (xi )i∈Z ∈ X to the sequence y = (yi )i∈Z given by yi =

1 if xi = xi+1 = 0 0 otherwise,

is well defined, continuous and surjective, ant that it commutes with the shift. 7.6 Let K be a finite discrete topological space with cardinality k. Given an integer n ≥ 1, let πn : K Z → K n be the map defined by

Exercises

137

πn (x) := (x0 , x1 , . . . , xn−1 ) for all x = (xi )i∈Z ∈ K Z . Let X ⊂ K Z be a non-empty subshift. For n ≥ 1, let γn (X ) denote the cardinality of the set πn (X ). (a) Show that the sequence (log γn (X ))n≥1 is subadditive. (b) Show that the limit log γn (X ) h(X ) := lim n→∞ n

(c) (d) (e)

(f) (g)

exists and satisfies 0 ≤ h(X ) ≤ log k. This limit is called the entropy of the subshift X . Show that h(X ) = h top (X, σ), where h top (X, σ) is the topological entropy of the dynamical system (X, σ) (cf. Exercise 6.11). Take K = {0, 1} and suppose that X ⊂ K Z is the subshift considered in Exercise 7.4. Show that h(X ) = 0. Take again K = {0, 1}. Show that the golden mean subshift X and the even subshift Y (cf. Exercise 7.5) have the same entropy h(X ) = h(Y ) = log φ, √ where φ := (1 + 5)/2 is the golden mean. Hint: check that γn+2 (X ) = γn+1 (X ) + γn (X ) and γn (Y ) = γn+1 (X ) − 1 for all n ≥ 1. Show that if Y ⊂ K Z is a subshift such that X ⊂ Y , then h(X ) ≤ h(Y ). Show that if X is the subshift of block type associated with a pair (q, B), where q ≥ 1 is an integer and B ⊂ K q has cardinality b, then h(X ) =

log b . q log k

(h) Show that if ρ is a real number such that 0 ≤ ρ ≤ log k, then there exists a subshift X ⊂ K Z such that h(X ) = ρ. Hint: adapt the ideas used in the proof of Theorem 7.6.2. 7.7 Let X be a compact metrizable space and T : X → X a homeomorphism. Let d be a metric on X defining the topology. One says that the homeomorphism T is expansive if there exists a real number δ > 0 satisfying the following condition: if two points x, y ∈ X are such that d(T n (x), T n (y)) ≤ δ for all n ∈ Z then x = y. (a) Show that the above definition does not depend on the choice of d. (b) Show that the following conditions are equivalent: (1) the homeomorphism T is expansive and dim(X ) = 0; (2) there exist a finite discrete topological space K and a subshift Y ⊂ K Z such that the dynamical systems (X, T ) and (Y, σ) are topologically conjugate. 7.8 Let K be a finite discrete topological space and let σ : K Z → K Z denote the shift map. One says that a map f : K Z → K Z is a cellular automaton if f is continuous and satisfies f ◦ σ = σ ◦ f .

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(a) Let f : K Z → K Z be a map. Show that f is a cellular automaton if and only if it satisfies the following condition: there exist an integer n ≥ 0 and a map μ : K 2n+1 → K such that, for all x = (xi )i∈Z ∈ K Z , one has f (x) = (yi )i∈Z , where yi = μ(xi−n , xi−n+1 , . . . , xi+n ) for all i ∈ Z. (b) Show that if f : K Z → K Z and g : K Z → K Z are cellular automata, then their composite map f ◦ g : K Z → K Z is also a cellular automaton. (c) Show that if f : K Z → K Z is a bijective cellular automaton, then its inverse map f −1 : K Z → K Z is also a cellular automaton. 7.9 Let K := [0, 1] denote the unit segment and σ : K Z → K Z the shift map. Let X denote the subset of K Z consisting of all x = (xi )i∈Z ∈ K Z such that xi + xi+1 = 1 for all i ∈ Z. (a) Show that X is a subshift of finite type of K Z . (b) Show that (X, σ) is topologically conjugate to the dynamical system (K , f ), where f : K → K is the map defined by f (t) = 1 − t for all t ∈ K . (c) Show that mdim(X, σ) = 0. 7.10 Let K ⊂ C denote the unit circle K := S1 = {z ∈ C : |z| = 1} and let σ : K Z → K Z be the shift map. Consider the subset X ⊂ K Z defined by 3 for all n ∈ Z}. X := {(z n )n∈Z ∈ K Z : z n2 = z n+1

(a) Show that X is a subshift of finite type. (b) Show that mdim(X, σ) = 0. Hint: observe that wn (X ), where wn : K Z → K n is defined by (7.3.1), is homeomorphic to S1 for all n ≥ 1 and then apply Theorem 7.3.5. 7.11 Let K be a compact Hausdorff space and let σ : K Z → K Z denote the shift map. Show that a closed subset X ⊂ K Z is a subshift of finite type if and only if it satisfies the following condition: there exists a finite subset ⊂ Z such that X = {x ∈ K Z | π (σ n (x)) ∈ π (X ) for all n ∈ Z} (here π : K Z → K denotes the canonical projection map).

Chapter 8

Applications of Mean Dimension to Embedding Problems

In this chapter, we prove the embedding theorem of Jaworski (Theorem 8.3.1) which asserts that every dynamical system (X, T ), where T is a homeomorphism without periodic points of a compact metrizable space X such that dim(X ) < ∞, embeds in the shift (RZ , σ). We also describe a family of counterexamples due to Lindenstrauss and Weiss showing that one cannot remove the hypothesis that X has finite topological dimension in the statement of Jaworski’s theorem, even if the dynamical system (X, T ) is assumed to be minimal.

8.1 Generalities Let X be a topological space and T : X → X a homeomorphism. The orbit of a point x ∈ X is the set OT (x) := {T n (x) | n ∈ Z}. The orbits of the points of X form a partition of X . One says that a point x ∈ X is a fixed point of T if T (x) = x. This amounts to saying that the orbit of x is reduced to the point x itself. The set Fix(T ) := {x ∈ X | T (x) = x} is a T -invariant subset of X . Note that Fix(T ) is closed in X if X is Hausdorff. One says that x ∈ X is a periodic point of T if the orbit of x is finite. The set Per(T ) consisting of all periodic points of T is a T -invariant subset of X and one has Per(T ) =

Per n (T ),

n≥1

where Per n (T ) := Fix(T n ) is the set consisting of all fixed points of T n . Observe that the set Per n (T ) is also T -invariant for each n. © Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_8

139

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8 Applications of Mean Dimension to Embedding Problems

Example 8.1.1 Let K be a topological space and σ : K Z → K Z the shift map. Then the set Per n (σ) consists of all sequences (xi ) ∈ K Z such that xi+n = xi for all i ∈ Z. The map h : Per n (σ) → K n defined by h(x) := (x0 , x1 , . . . , xn−1 ) is a homeomorphism. Moreover, if τn denotes the cyclic permutation of coordinates on K n , i.e., the map (x0 , x1 , . . . , xn−1 ) → (x1 , x2 , . . . , x0 ), then h conjugates the dynamical systems (Per n (σ), σ) and (K n , τn ). In particular, Per 1 (σ) = Fix(σ), which consists of all constant sequences in K Z , is homeomorphic to K . One says that the dynamical system (X, T ) is topologically transitive if there exists a point in X whose orbit is dense in X . One says that the dynamical system (X, T ) is minimal, or that the homeomorphism T is minimal, if the orbit of every point of X is dense in X . This amounts to saying that the only closed T -invariant subsets of X are X and the empty set. Remark 8.1.2 Let X be an accessible topological space and T : X → X a homeomorphism admitting a periodic point x. Then the dynamical system (X, T ) is minimal if and only if X is reduced to the orbit of x. Let X and Y be topological spaces. Let T : X → X and S : Y → Y be homeomorphisms. One says that the dynamical system (X, T ) embeds in the dynamical system (Y, S) if there exists a topological embedding f : X −→ Y such that f ◦ T = S ◦ f . One then says that f is an embedding of the dynamical system (X, T ) in the dynamical system (Y, S). Note that the system (X, T ) embeds in (Y, S) if and only if there exists a S-invariant subset Z ⊂ Y such that (X, T ) is topologically conjugate to (Z , S). Clearly every embedding of (X, T ) in (Y, S) induces an embedding of (Per n (T ), T ) in (Per n (S), S) for every integer n ≥ 1. In particular, a necessary condition for the existence of an embedding of (X, T ) in (Y, S) is that the space Per n (T ) is embeddable in Per n (S) for every n ≥ 1. Example 8.1.3 Let S2 = {(x1 , x2 , x3 ) ∈ R3 | x12 + x22 + x32 = 1} be the unit sphere in R3 . Consider the equatorial symmetry τ : S2 → S2 , i.e., the homeomorphism of S2 defined by τ (x1 , x2 , x3 ) = (x1 , x2 , −x3 ). Then the dynamical system (S2 , τ ) cannot be embedded into the shift (RZ , σ) because the set of fixed points of τ , which is a circle, cannot be embedded into the set of fixed points of σ, which is homeomorphic to R. Note that the dynamical system (S2 , τ ) embeds into the shift ((R2 )Z , σ). Indeed, one easily verifies that the map f : S2 → (R2 )Z that sends each point x = (x1 , x2 , x3 ) ∈ S2 to the sequence u = (u i ) ∈ (R2 )Z defined by ui =

(x1 + x3 , x2 ) if i is even, (x1 − x3 , x2 ) if i is odd

yields an embedding of (S2 , τ ) into ((R2 )Z , σ).

8.2 Embeddings in Shifts

141

8.2 Embeddings in Shifts Proposition 8.2.1 Let K and L be topological spaces. Then the shift (L Z , σ L ) embeds in the shift (K Z , σ K ) if and only if the space L embeds in K . Proof If the shift (L Z , σ L ) embeds in the shift (K Z , σ K ), then Fix(σ L ) embeds in Fix(σ K ). As Fix(σ L ) (resp. Fix(σ K )) is homeomorphic to L (resp. K ), this implies that L is embeddable in K . Conversely, suppose that ϕ : L −→ K is an embedding. Then the map f : L Z → Z K , defined by f (x) = (ϕ(xi ))i∈Z for every x = (xi )i∈Z ∈ L Z is clearly an embedding of (L Z , σ L ) in (K Z , σ K ). Example 8.2.2 Let X and Y be finite sets equipped with their discrete topology. Then X embeds in Y if and only if #(X ) ≤ #(Y ). Consequently, the shift (X Z , σ X ) embeds in the shift (Y Z , σY ) if and only if #(X ) ≤ #(Y ). Proposition 8.2.3 Let X be a topological space and T : X → X a homeomorphism. Suppose that K is a topological space such that X embeds in K . Then the dynamical system (X, T ) embeds in the shift (K Z , σ). Proof Let ϕ : X −→ K be a topological embedding. Consider the map f : X → K Z that sends each point x ∈ X to the sequence (u i ) ∈ K Z defined by u i = ϕ(T i (x)) for all i ∈ Z. Clearly f satisfies f ◦ T = σ ◦ f . On the other hand, f induces a bijective map from X onto f (X ) whose inverse is the map π : f (X ) → X given by (u i ) → ϕ−1 (u 0 ). As the maps f and π are obviously continuous, we deduce that f is an embedding of (X, T ) in (K Z , σ). Corollary 8.2.4 Let X be a compact metrizable space and T : X → X a homeomorphism. Let K := [0, 1]N denote the Hilbert cube. Then the dynamical system (X, T ) embeds in the shift (K Z , σ). Proof By Proposition 2.4.18, every compact metrizable space embeds in the Hilbert cube. Corollary 8.2.5 Let X be a compact metrizable space such that dim(X ) < ∞ and T : X → X a homeomorphism. Then the dynamical system (X, T ) embeds in the shift ((Rn )Z , σ) for n = 2 dim(X ) + 1. Proof The space X embeds in K := Rn by Corollary 4.7.6.

Remark 8.2.6 Let K be a compact metrizable space with dim(K ) = ∞ (e.g., the Hilbert cube [0, 1]N ). Let T denote the shift map on the compact metrizable space X := K Z . Then there is no integer n such that the dynamical system (X, T ) embeds in the shift ((Rn )Z , σ). Indeed, the set of fixed points of the shift σ on (Rn )Z is homeomorphic to Rn while the set of fixed points of T , which is homeomorphic to K , cannot be embedded in Rn .

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8 Applications of Mean Dimension to Embedding Problems

Proposition 8.2.7 Let X be a compact space and K a Hausdorff space. Let T : X → X be a homeomorphism. Then the following conditions are equivalent: (a) the dynamical system (X, T ) embeds in the shift (K Z , σ); (b) there exists a continuous map f : X → K satisfying the following condition: given any two distinct points x and y in X , there is an integer i ∈ Z such that f (T i (x)) = f (T i (y)). Proof Suppose (a). Then there exists an embedding ϕ : X → K Z such that ϕ ◦ T = σ ◦ ϕ. Let x and y be distinct points of X . The injectivity of ϕ implies the existence of an integer i ∈ Z such that the terms of rank i of the sequences ϕ(x) and ϕ(y) are distinct. Therefore, we have that π ◦ σ i ◦ ϕ(x) = π ◦ σ i ◦ ϕ(y), where π : K Z → K denotes the projection on the 0-factor. As σ i ◦ ϕ = ϕ ◦ T i , it follows that the map f = π ◦ ϕ : X → K satisfies f (T i (x)) = f (T i (y)). This shows that (a) implies (b). Conversely, suppose that f : X → K is a map as in (b). Consider the map ϕ : X → K Z that sends each point x ∈ X to the sequence (u i ) ∈ K Z defined by u i = f (T i (x)) for all i ∈ Z. Clearly ϕ is continuous and satisfies σ ◦ ϕ = ϕ ◦ T . Our hypothesis on f implies that ϕ is injective. As X is compact and K Z is Hausdorff, we deduce that ϕ induces a homeomorphism from X onto ϕ(X ). This shows that (b) implies (a). Example 8.2.8 Take as X the unit circle S1 = {(x1 , x2 ) ∈ R2 | x12 + x22 = 1}. Let α be a real number that is not an integer multiple of π and let T : X → X denote the rotation of angle α, i.e., the map given by T (x) = (x1 cos α − x2 sin α, x1 sin α + x2 cos α) for all x = (x1 , x2 ) ∈ X . Let σ denote the shift map on RZ . Consider the vertical projection f : X → R, i.e., the map defined by f (x1 , x2 ) = x1 . If x and y are distinct points in X that lie on the same vertical, then the angle between the horizontal and the line passing through the points T (x) and T (y) is π/2 + α, so that f (T (x)) = f (T (y)). Thus, every pair of distinct points x, y ∈ X satisfy f (x) = f (y) or f (T (x)) = f (T (y)). By applying Proposition 8.2.7, we deduce that the dynamical system (X, T ) embeds in the shift (RZ , σ).

8.3 Jaworski’s Embedding Theorem

143

8.3 Jaworski’s Embedding Theorem Let X be a compact metrizable space with dim(X ) < ∞ and T : X → X a homeomorphism. By Corollary 8.2.5, there exists an integer n ≥ 1 such that the dynamical system (X, T ) embeds in the shift ((Rn )Z , σ). A natural question that immediately arises is to determine the smallest integer n such that (X, T ) can be embedded in the shift ((Rn )Z , σ). Let us denote this integer n 0 (X, T ). For example, if τ is the equatorial symmetry on the sphere S2 then n 0 (S2 , τ ) = 2 (see Example 8.1.3). The following theorem shows that one always has n 0 (X, T ) = 1 if T has no periodic points. Theorem 8.3.1 (Jaworski’s embedding theorem) Let X be a compact metrizable space with dim(X ) < ∞ and let T : X → X be a homeomorphism without periodic points. Then the dynamical system (X, T ) embeds in the shift (RZ , σ). Let (X, T ) be a dynamical system satisfying the hypotheses of Theorem 8.3.1. We shall use the following notation. The vector space consisting of all continuous maps f : X → R is denoted by C(X ). We equip C(X ) with the sup-norm · ∞ given by f ∞ = supx∈X | f (x)|. Let denote the complement of the diagonal in X × X , that is, the open subset of X × X defined by := {(x, y) ∈ X × X | x = y}. For each subset K ⊂ , we denote by D(K ) the set consisting of all maps f ∈ C(X ) satisfying the following property: for every (x, y) ∈ K , there exists i ∈ Z such that f (T i (x)) = f (T i (y)). Let us start by establishing some auxiliary results. Lemma 8.3.2 Let K be a compact subset of . Then D(K ) is an open subset of C(X ). Proof Let f ∈ D(K ). Then the map H : X × X → R defined by H (x, y) := sup | f (T i (x)) − f (T i (y))| i∈Z

satisfies H (x, y) > 0 for all (x, y) ∈ K . The map H is lower semi-continuous since it is the supremum of a family of continuous maps. As K is compact, the restriction of H to K attains its infimum. It follows that there exists a real number δ > 0 such that H (x, y) ≥ δ for all (x, y) ∈ K . Consider now a map g ∈ C(X ) such that f − g∞ ≤ δ/4. Let (x, y) ∈ K . Using the triangle inequality, we obtain, for every i ∈ Z, |g(T i (x)) − g(T i (y))| ≥ | f (T i (x)) − f (T i (y))| − | f (T i (x)) − g(T i (x))| − | f (T i (y)) − g(T i (y))| δ ≥ | f (T i (x)) − f (T i (y))| − . 2

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8 Applications of Mean Dimension to Embedding Problems

We deduce that sup |g(T i (x)) − g(T i (y))| ≥ H (x, y) − i∈Z

δ δ ≥ > 0. 2 2

It follows that g ∈ D(K ). Consequently, the set D(K ) is open in C(X ).

Lemma 8.3.3 Let (x, y) ∈ and let m be a positive integer. Then there exist integers i 1 , . . . , i m ∈ Z such that the points T i1 (x), . . . , T im (x), T i1 (y), . . . , T im (y) are pairwise distinct. Proof By our hypotheses, the homeomorphism T has no periodic points. This means that, for every z ∈ X , the map i → T i (z) yields a bijection from Z onto the orbit of z. If x and y are not in the same orbit, we can take as i 1 , . . . , i m arbitrary pairwise distinct integers. Otherwise, we have that y = T n (x) for some integer n = 0. Then we can construct a sequence i 1 , . . . , i m by induction on m in the following way. For m = 1, we can take any i 1 ∈ Z. Suppose that i 1 , . . . , i m have already been constructed. Then we can take as i m+1 any integer that does not belong to the set {i 1 , . . . , i m } ∪ {i 1 + n, . . . , i m + n} ∪ {i 1 − n, . . . , i m − n}. Lemma 8.3.4 Let (x0 , y0 ) ∈ . Then there exists a compact neighborhood K of (x0 , y0 ) in such that D(K ) is dense in C(X ). Proof Let m be an integer such that m ≥ 2 dim(X ) + 1. By applying Lemma 8.3.3, we can find integers i 1 , . . . , i m ∈ Z such that the points T i1 (x0 ), . . . , T im (x0 ), T i1 (y0 ), . . . , T im (y0 ) are pairwise distinct. As X is compact and Hausdorff, there exist a compact neighborhood V of x0 in X and a compact neighborhood W of y0 in X such that the sets T i1 (V ), . . . , T im (V ), T i1 (W ), . . . , T im (W ) are pairwise disjoint. Observe that V ∩ W = ∅ and that the set K := V × W is a compact neighborhood of (x0 , y0 ) in . Let us show that D(K ) is dense in C(X ). Suppose that f ∈ C(X ) and let ε > 0. Consider the map ϕ : V ∪W → Rm defined by ϕ(x) := ( f (T i1 (x)), . . . , f (T im (x))) for all x ∈ V ∪ W . As m ≥ 2 dim(X ) + 1 ≥ 2 dim(V ∪ W ) + 1 by Proposition 1.2.1, we deduce from Theorem 4.7.5 that there exists an embedding ψ : V ∪ W −→ Rm such that ϕ(x) − ψ(x) ≤ ε for all x ∈ V ∪ W, (8.3.1)

8.3 Jaworski’s Embedding Theorem

145

where · denotes the Euclidean norm on Rm . Write ψ = (ψ1 , . . . , ψm ) and Z := T i1 (V ) ∪ · · · ∪ T im (V ) ∪ T i1 (W ) ∪ · · · ∪ T im (W ). Consider now the map h : Z → R defined by h(z) := ψk (T −ik (z)) for all z ∈ T ik (V ) ∪ T ik (W ) (1 ≤ k ≤ m). Clearly h is well defined and continuous. Moreover, for every z ∈ T ik (V ) ∪ T ik (W ), we have, by setting x := T −ik (z) and using (8.3.1), |h(z) − f (z)| = |ψk (x) − f (T ik (x))| ≤ ψ(x) − ϕ(x) ≤ ε. It follows that |h(z) − f (z)| ≤ ε for all z ∈ Z . By Corollary 4.1.5, we can extend h to a continuous map g : X → R such that g − f ∞ ≤ ε. Let us show that g ∈ D(K ). Let (x, y) ∈ K = V × W . As ψ is injective, there exists k ∈ {1, . . . , m} such that ψk (x) = ψk (y). Since g(T ik (x)) = h(T ik (x)) = ψk (x) and g(T ik (y)) = h(T ik (y)) = ψk (y), we deduce that g(T ik (x)) = g(T ik (y)). Thus, we have that g ∈ D(K ). This shows that D(K ) is dense in C(X ). Proof of Theorem 8.3.1 By virtue of Proposition 8.2.7, it is enough to prove that D() is not empty. It follows from Lemma 8.3.2 and Lemma 8.3.4 that, for every (x, y) ∈ , we can find a compact neighborhood K of (x, y) in such that D(K ) is a dense open subset of C(X ). The space ⊂ X × X is a Lindelöf space since every subset of a compact metrizable space is Lindelöf by Corollary 2.4.14 and Proposition 2.4.18. Therefore we can cover by a sequence (K n )n∈N of compact subsets such that D(K n ) is a dense open subset of C(X ) for every n.As C(X ) is complete, we can apply Baire’s theorem and conclude that D() = n∈N D(K n ) is dense in C(X ) and hence non-empty. Corollary 8.3.5 Let X be a compact metrizable space with dim(X ) < ∞ and let T : X → X be a minimal homeomorphism. Then the dynamical system (X, T ) embeds in the shift (RZ , σ). Proof If T has no periodic points, this follows from Theorem 8.3.1. On the other hand, if x ∈ X is a periodic point of T , then the minimality of T implies that X is reduced to the orbit of x (see Remark 8.1.2). Consequently, the set X is finite. As every finite discrete space can be topologically embedded in the real line, we conclude that the dynamical system (X, T ) embeds in (RZ , σ) by applying Proposition 8.2.3.

8.4 The Lindenstrauss-Weiss Counterexamples The goal of this section is to establish the following result which shows that the finiteness hypothesis on the topological dimension of X cannot be removed from the statement of Corollary 8.3.5.

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8 Applications of Mean Dimension to Embedding Problems

Theorem 8.4.1 There exists a minimal dynamical system (X, T ), where X is a compact metrizable space and T : X → X a homeomorphism, that cannot be embedded in the shift (RZ , σ). Remark 8.4.2 A dynamical system (X, T ) that satisfies the conditions of Theorem 8.4.1 has no periodic points (see the proof of Corollary 8.3.5). Consequently, Theorem 8.4.1 also shows that the finiteness hypothesis on the topological dimension of X cannot be removed from the statement of Theorem 8.3.1. Theorem 8.4.1 is an immediate consequence of the following result. Theorem 8.4.3 Let N ≥ 2 be an integer and let K := [0, 1] N denote the n-dimensional cube. Then there exists a subshift X ⊂ K Z such that the dynamical system (X, σ) is minimal and satisfies mdim(X, σ) > 1. Indeed, if X is a compact metrizable space and T : X → X a homeomorphism such that the dynamical system (X, T ) embeds in (RZ , σ), then (X, T ) embeds in ([0, 1]Z , σ) (since R embeds in [0, 1]), and hence mdim(X, T ) ≤ mdim([0, 1]Z , σ) = 1 by Corollary 6.4.7 and Theorem 7.2.1. The remainder of this section is devoted to the proof of Theorem 8.4.3. Let us first establish some auxiliary results. Lemma 8.4.4 Let K be a topological space, q ∈ N, and B a separable closed subset of K q . Let X ⊂ K Z denote the subshift of block-type associated with (q, B). Then the dynamical system (X, σ) is topologically transitive. Proof Let A be a countable dense subset of B. As the disjoint union k≥1 Ak is countable, there exists a sequence (u n )n∈Z of elements of A satisfying the following property: for every integer k ≥ 1 and every (a1 , . . . , ak ) ∈ Ak , there exists n 0 ∈ Z such that (u n 0 , . . . , u n 0 +k−1 ) = (a1 , . . . , ak ). Consider now the sequence x = (xi )i∈Z ∈ X defined by (xqn , . . . , xqn+q−1 ) = u n for all n ∈ Z. Then, it follows from our construction that, given any non-empty open subset U of X , there exists m ∈ Z such that σ m (x) ∈ U . Consequently, the orbit of x is dense in X . This shows that the dynamical system (X, T ) is topologically transitive. Let (X, d) be a metric space. Given a real number ε > 0, one says that a subset Y ⊂ X is ε-dense in X if, for every x ∈ X , there exists y ∈ Y such that d(x, y) ≤ ε. Lemma 8.4.5 Let (E, d) be a metric space and T : E → E a homeomorphism. Let (X n )n∈ N be a sequence of subsets of E such that T (X n ) = X n for all n ∈ N. Let X := n∈N X n . Suppose that there exists a sequence (εn )n∈N of positive real numbers converging to 0 such that, for all n ∈ N and x ∈ X n , the T -orbit of x is εn -dense in X n . Then the dynamical system (X, T ) is minimal. Proof Let x and y be points in X . Then, for every n ∈ N, we have that x, y ∈ X n . Therefore, we can find an integer kn ∈ Z such that d(T kn (x), y) ≤ εn . As the sequence (εn ) converges to 0, this implies that the orbit of x is dense in X .

8.4 The Lindenstrauss-Weiss Counterexamples

147

Lemma 8.4.6 Let (X, d) be a compact metric space and T : X → X a homeomorphism. Suppose that y is a point of X with dense T -orbit and let ε > 0. Then there is an integer m ≥ 0 such that, for every x ∈ X , there exists an integer k ∈ {−m, . . . , m} such that d(x, T k (y)) < ε. Proof As the orbit of y is dense, we can find for every x ∈ X , n(x) ∈ Z such that d(x, T n(x) (y)) < ε. By continuity of the metric, there exists a neighborhood Vx of x such that d(z, σ n(x) (y)) < ε for all z ∈ Vx . The compactness of X implies that there exists a finite subset A ⊂ X such that the sets Vx , where x ∈ A, cover X . Then the integer m := max x∈A |n(x)| has the required property. The upper-density δ(J ) ∈ [0, 1] of a subset J ⊂ N is defined by δ(J ) := lim sup n→∞

#(J ∩ {0, 1, . . . , n − 1}) . n

Example 8.4.7 If J is a finite subset of N, then δ(J ) = 0. Example 8.4.8 If J := {1, 3, 5, . . . } is the subset of N consisting of all positive odd 1 numbers, then δ(J ) = . 2 In what follows, we fix some integer N ≥ 0 and K := [0, 1] N denotes the N dimensional cube. For A ⊂ Z, we denote by π A : K Z → K A the projection map, i.e., the map given by π A (x) := (xi )i∈A ∈ K A for all x = (xi )i∈Z ∈ K Z . We denote by d K the metric induced on K ⊂ R N by the norm · ∞ and we equip K Z with the metric d defined by Formula (7.1.1). Lemma 8.4.9 Let X ⊂ K Z be a subshift. Suppose that there exist an element x = (x i )i∈Z ∈ X and a subset J ⊂ N satisfying the following condition: if x = (xi )i∈Z ∈ K Z is such that πZ\J (x) = πZ\J (x), then one has x ∈ X . Then X satisfies mdim(X, σ) ≥ N δ(J ). Proof (cf. the proof of Theorem 7.2.1 and that of Lemma 7.6.1) Recall that d K (x0 , y0 ) ≤ d(x, y) for all x = (xi ), y = (yi ) ∈ K Z . Consider, for every n ≥ 1, the metric dn on X defined by dn (x, y) :=

max d(σ k (x), σ k (y)).

0≤k≤n−1

Define the subset Jn ⊂ N by Jn := J ∩ {0, 1, . . . , n − 1},

(8.4.1)

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8 Applications of Mean Dimension to Embedding Problems

and denote by ρn the metric induced on K Jn = [0, 1] N #(Jn ) ⊂ R N #(Jn ) by the norm · ∞ . Consider the topological embedding ϕn : K Jn −→ K Z that sends each u = (u j ) j∈Jn ∈ K Jn to the sequence x = (xi )i∈Z given by xi :=

u i if i ∈ Jn , x i if i ∈ Z\Jn .

By our hypothesis on x, we have that ϕn (u) ∈ X for all u ∈ K Jn . On the other hand, Inequality (8.4.1) implies that ρn (u, v) ≤ dn (ϕn (u), ϕn (v)) for all u, v ∈ K Jn . By applying Proposition 4.6.3, we deduce that dimε (X, dn ) ≥ dimε (K Jn , ρn ), for every ε > 0. As dimε (K Jn , ρn ) = N #(Jn ) for all ε ≤ 1 by Proposition 4.6.5, it follows that dimε (X, dn ) ≥ N #(Jn ), for all ε ≤ 1. Since, by definition, dimε (X, dn ) , n→∞ n

mdimε (X, d, σ) = lim we deduce that

mdimε (X, d, σ) ≥ N δ(J ) for all ε ≤ 1. By letting ε tend to 0, we finally get mdim(X, σ) = lim mdimε (X, d, σ) ≥ N δ(J ). ε→0

Lemma 8.4.10 Let (εn )n∈N be a sequence of positive real numbers converging to 0. Let X (n) ⊂ K Z , n ∈ N, be a sequence of subshifts such that there exists, for each n ∈ N, an element x (n) ∈ X (n) and a subset I (n) ⊂ Z satisfying, for all n ∈ N, the following properties: (P1) (P2) (P3) (P4) (P5)

the orbit of every point of X (n) is εn -dense in X (n) , X (n+1) ⊂ X (n) , I (n+1) ⊂ I (n) , πZ\I (n) (x (n+1) ) = πZ\I (n) (x (n) ), if x ∈ K Z satisfies πZ\I (n) (x) = πZ\I (n) (x (n) ), then one has x ∈ X (n) .

8.4 The Lindenstrauss-Weiss Counterexamples

149

Let X ⊂ K Z be the subshift defined by

X :=

X (n) .

n∈N

Then the dynamical system (X, σ) is minimal and one has mdim(X, σ) ≥ N δ(J ), where δ(J ) denotes the upper density of the subset J ⊂ N defined by J := N ∩

I

(n)

.

n∈N

Proof Since the sequence (εn )n∈N converges to 0, Property (P1) implies that the dynamical system (X, σ) is minimal by Lemma 8.4.5. By compactness of K Z , we can extract from the sequence (x (n) ) a subsequence that converges to some element x ∈ K Z . Since each X (n) is closed in K Z , we deduce from (P2) that x ∈ X . Moreover, Properties (P3) and (P4) imply that πZ\I (n) (x) = πZ\I (n) (x (n) ) for all n ∈ N. By using (P5), we deduce that every element x ∈ K Z such that πZ\J (x) = πZ\J (x) belongs to X . It follows that mdim(X, σ) ≥ N δ(J ) by Lemma 8.4.9. Proof of Theorem 8.4.3 Let us first choose a sequence of positive real numbers (εn )n∈N converging to 0 such that ε0 = diam(K Z ) = 3. We also fix a sequence (an )n∈N of positive integers (a condition on the sequence (an )n∈N will be added at the end of the proof). Let us construct, by induction on n ∈ N, a sequence of pairs (qn , Bn ), where qn is a positive integer and Bn is a closed subset of K qn , and a sequence of subsets (n) I (n) ⊂ Z such that the subsets X 0 ⊂ X (n) ⊂ K Z defined by (n)

X 0 := {x = (xi )i∈Z ∈ K Z | (xi , . . . , xi+qn −1 ) ∈ Bn for all i ∈ qn Z}, and

X (n) :=

(n)

σ k (X 0 )

k∈Z

satisfy the following properties: (Q1) the orbit of every point of X (n) is εn -dense in X (n) , (n) (Q2) πZ\I (n) (x) = πZ\I (n) (x ) for all x, x ∈ X 0 , (n+1)

(n)

⊂ X 0 (and hence X (n+1) ⊂ X (n) ), (Q3) X 0 (n+1) ⊂ I (n) (Q4) I for all n ∈ N.

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8 Applications of Mean Dimension to Embedding Problems (0)

We start by taking q0 := 1, B0 := K , and I (0) := Z. This gives us X 0 = X (0) = K Z so that Properties (Q1) and (Q2) are trivially satisfied for n = 0. Suppose now that qn , Bn , and I (n) have already been constructed for some n ≥ 0. Then we construct qn+1 , Bn+1 , and I (n+1) in the following way. Since the space Bn is a subspace of R N qn , it is separable by Corollary 2.4.14. As X (n) is the subshift of block-type associated with the pair (qn , Bn ), we deduce from Lemma 8.4.4 that the dynamical system (X (n) , σ) is topologically transitive. This means that we can find an element y ∈ X (n) whose orbit is dense in X (n) . After possibly replacing y by a point in its orbit, we can assume that y ∈ X 0(n) . Choose an integer rn+1 ≥ 1 large enough so that, for all x, x ∈ K Z , π{−rn+1 ,...,rn+1 } (x) = π{−rn+1 ,...,rn+1 } (x ) ⇒ d(x, x ) ≤

εn+1 . 2

By Lemma 8.4.6, we can find an integer L n+1 ≥ rn+1 such that, for every x ∈ X (n) , there exists k ∈ {−L n+1 + rn+1 , . . . , L n+1 − rn+1 } satisfying d(x, σ k (y)) ≤

εn+1 . 2

We can assume in addition that the integer L n+1 is a multiple of qn . Put qn+1 := (an+1 + 1)2L n+1 . Let Bn+1 denote the closed subset of K qn+1 consisting of all b = (b1 , . . . , bqn+1 ) ∈ Bn × · · · × Bn ⊂ K qn+1 ,

qn+1 qn

times

ending by the finite sequence (y−L n+1 , . . . , y L n+1 −1 ), that is, such that (bqn+1 −2L n+1 +1 , . . . , bqn+1 ) = (y−L n+1 , . . . , y L n+1 −1 ). Consider now the subsets X 0(n+1) ⊂ X (n+1) defined by (qn+1 , Bn+1 ). Then, we have (n+1) (n) ⊂ X 0 (and hence X (n+1) ⊂ X (n) ) since that X 0 Bn+1 ⊂ Bn × · · · × Bn . As the sequence (y−L n+1 , . . . , y L n+1 −1 ) appear in every element of X (n+1) , it follows from our choices of rn+1 and L n+1 that the orbit of every point of X (n+1) is εn+1 -dense in X (n) and hence in X (n+1) . Consequently, Property (Q1) is satisfied at rank n + 1. Denote by Rn+1 the set consisting of all integers i ∈ Z such that there exists k ∈ {0, . . . , qn+1 − 2L n+1 − 1}

8.4 The Lindenstrauss-Weiss Counterexamples

151

with i ≡ k mod qn+1 . Let I (n+1) ⊂ I (n) be the subset defined by I (n+1) := I (n) ∩ Rn+1 . Suppose that x, x ∈ X 0(n+1) . Then we have that πZ\Rn+1 (x) = πZ\Rn+1 (x ) by (n+1) definition of Bn+1 . Moreover, we have that πZ\I (n) (x) = πZ\I (n) (x ) since X 0 ⊂ (n)

X 0 and Property (Q2) is satisfied at rank n by our induction hypothesis. As Z\I (n+1) = (Z\I (n) ) ∪ (Z\Rn+1 ), we deduce that πZ\I (n+1) (x) = πZ\I (n+1) (x ). This shows that (Q2) is satisfied at rank n + 1. This completes the construction by induction. (n) Let us choose an arbitrary element x (n) ∈ X 0 for each n ∈ N. Then we have that (Q5) πZ\I (n) (x (n+1) ) = πZ\I (n) (x (n) ), by (Q2) and (Q3). Let us show by induction on n the following property: (n)

(Q6) if x ∈ K Z satisfies πZ\I (n) (x) = πZ\I (n) (x (n) ) then x ∈ X 0 . For n = 0, there is nothing to prove. Suppose now that the statement is true at rank n and let x ∈ K Z such that πZ\I (n+1) (x) = πZ\I (n+1) (x (n+1) ). By using (Q4) and (Q5), we obtain πZ\I (n) (x) = πZ\I (n) (x (n) ). (n)

This implies x ∈ X 0 by our induction hypothesis. On the other hand, as Z\Rn+1 ⊂ Z\I (n+1) , we also have that πZ\Rn+1 (x) = πZ\Rn+1 (x (n+1) ). (n+1)

Thus, we finally get x ∈ X 0 Consider now the subshift

, which completes the proof by induction of (Q6). X (n) . X := n∈N

It follows from Properties (Q1), (Q3), (Q4), (Q5), and (Q6) that X (n) , x (n) , and I (n) satisfy the hypotheses of Lemma 8.4.10. Consequently, the dynamical system (X, σ) is minimal and (8.4.2) mdim(X, σ) ≥ N δ(J ),

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8 Applications of Mean Dimension to Embedding Problems

where δ(J ) ∈ [0, 1] is the upper-density of the subset J ⊂ N defined by J := N ∩

I

(n)

=N∩

n∈N

∞

Rn .

n=1

Recall that, by definition, δ(J ) = lim sup n→∞

#(Jn ) , n

where Jn := J ∩ {0, . . . , n − 1}. For every m ≥ n + 1, we have that qn ≤ qm − 2L m and hence {0, . . . , qn − 1} ⊂ Rm . It follows that n Jqn = Rk . k=1

We deduce that n n

#(Jqn ) 1 2ak L k ak = n = = −1 qn 2ak L k + 2L k ak + 1 k=1 (1 + ak ) k=1

k=1

for all n ≥ 1. This implies 1

δ(J ) ≥ ∞

−1 n=1 (1 + an )

.

(8.4.3)

Let us choose now the sequence (an ) so that ∞

(1 + an−1 ) < 2.

(8.4.4)

n=1

We can take for example an = 2n+1 since the inequality log(1 + x) ≤ x (x ≥ 0) implies that ∞ ∞

√ −n−1 −n−1 (1 + 2 ) ≤ exp 2 = e < 2. n=1

n=1

Finally, Inequalities (8.4.2)–(8.4.4) give us mdim(X, σ) > N /2. This implies mdim(X, σ) > 1 since N ≥ 2.

Notes

153

Notes Theorem 8.3.1 was obtained by Jaworski in his Ph.D. thesis [51, Th. IV.1] (see [11, p. 194]). A homeomorphism T of a topological space X generates a continuous action of the additive group Z on X , given by the map Z × X → X that sends each pair (n, x) ∈ Z × X to the point T n (x). A continuous action of the additive group R is called a flow. More precisely, a flow on a topological space X consists of a family ϕ = (ϕt )t∈R of homeomorphisms of X such that the map (t, x) → ϕt (x) is continuous and ϕt ◦ ϕs = ϕt+s for all t, s ∈ R. A point x ∈ X is a fixed point of the flow (X, ϕ) if ϕt (x) = x for all t ∈ R. One says that the flow (X, ϕ) embeds in the flow (Y, ψ) if there exists a topological embedding h : X −→ Y such that h ◦ ϕt = ψt ◦ h for all t ∈ R. Denote by C(R) the set consisting of all continuous maps f : R → R and equip C(R) with the topology of uniform convergence on compact subsets of R. Consider the flow λ = (λt )t∈R on C(R) defined by λt ( f )(u) := f (u + t) for all f ∈ C(R) and t, u ∈ R. The flow (C(R), λ) is a continuous version of the shift (RZ , σ). Theorem 8.3.1 is analogous to a theorem of Bebutoff [13] (see also [53, 55], [11, p. 184]) which asserts that every flow without fixed points (X, ϕ), where X is a compact metrizable space, embeds in the flow (C(R), λ). Note however that, in contrast with Theorem 8.3.1, there is no hypothesis about the topological dimension of X in the statement of Bebutoff’s theorem. The set of fixed points of the flow (C(R), λ) is the set of constant functions and is therefore homeomorphic to the real line R. It follows that a necessary condition for a flow (X, ϕ) to be embeddable in the flow (C(R), λ) is that the set of fixed points of (X, ϕ) is homeomorphic to a subset of R. By a result of Kakutani [53], which extends Bebutoff’s theorem, it turns out that this condition is also sufficient for flows on compact metrizable spaces. The construction of the counterexamples described in Sect. 8.4 is due to Lindenstrauss and Weiss (see [74, Proposition 3.5]). Let T be a homeomorphism of a compact metrizable space X . In [72, Th. 5.1], Lindenstrauss proved that if (X, T ) is minimal and mdim(X, T ) < d/36 for some integer d ≥ 1, then the dynamical system (X, T ) can be embedded in the shift ((Rd )Z , σ). On the other hand, Lindenstrauss and Tsukamoto [73] constructed, for any integer d ≥ 1, a compact metrizable space X admitting a homeomorphism T such that the dynamical system (X, T ) is minimal and satisfies mdim(X, T ) = d/2 but cannot be embedded in the shift ((Rd )Z , σ). For additional results related to Jaworski’s theorem and the question of the embeddability of dynamical systems in the shift on (Rd )Z , see also [45, 46].

Exercises 8.1 Let K be a topological space and let σ denote the shift map on K Z . Show that the set of periodic points of σ is dense in K Z . 8.2 Let K be an accessible separable space with more than one point. Show that the shift (K Z , σ) is topologically transitive but not minimal.

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8 Applications of Mean Dimension to Embedding Problems

8.3 Let K be a topological space and let σ : K Z → K Z denote the shift map on K Z . Let q be a positive integer and B a closed subset of K q . Let X ⊂ K Z denote the subshift of block-type associated with (q, B). Show that the set of periodic points of the dynamical system (X, σ) is dense in X . 8.4 (Adding machines). Let (an )n∈N be a sequence of positive integers. Consider the product space

{0, 1, . . . , an − 1} X := n∈N

and the map T : X → X defined in the following way. If x = (xn )n∈N ∈ X with xn = an − 1 for all n ∈ N then we take T (x) := (yn )n∈N , where yn = 0 for all n ∈ N. Otherwise, there is a largest integer n 0 ∈ N such that xn = an −1 for all n ≤ n 0 , and we take T (x) := (yn )n∈N , where yn = 0 for all n ≤ n 0 , yn 0 +1 = xn 0 +1 + 1, and yn = xn for all n ≥ n 0 + 2. (a) Show that T is a homeomorphism. (b) Show that the dynamical system (X, T ) is minimal. 8.5 Let X be a topological space and T : X → X a homeomorphism. Show that the dynamical system (X, T ) is minimal if and only if every non-empty open subset U ⊂ X satisfies n∈Z T n (U ) = X . 8.6 Let X be a compact space and T : X → X a homeomorphism. Show that the dynamical system (X, T ) is minimal if and only if, for every non-empty open subset U ⊂ X , there exists n ∈ N such that nk=−n T k (U ) = X . 8.7 Let X be a non-empty compact Hausdorff space and T : X → X a homeomorphism. Show that there exists a non-empty closed subset Y ⊂ X with T (Y ) = Y such that the dynamical system (Y, T ) is minimal. Hint: use Zorn’s lemma. 8.8 Show that every closed subgroup G of R such that {0} = G = R is infinite cyclic. Deduce that if T : S1 → S1 is a rotation of angle θ with θ/π irrational, then the dynamical system (S1 , T ) is minimal. 8.9 Let S be a subset of Z. One says that S is syndetic if there exists a finite subset F ⊂ Z such that S + F = Z. Show that the following conditions are equivalent: (1) S is syndetic; (2) there exists an integer N ≥ 1 such that S is N -dense in Z; (3) there exists an integer k ≥ 1 such that one has {i, i + 1, . . . , i + k − 1} ∩ S = ∅ for all i ∈ Z; (4) S has bounded gaps, i.e., there exists an integer L ≥ 1 such that every subset of Z\S consisting of consecutive integers has cardinality at most L. 8.10 Let X be a topological space equipped with a homeomorphism T : X → X . A point x ∈ X is called almost-periodic if, for every neighborhood U of x, the set consisting of all n ∈ Z such that T n (x) ∈ U is a syndetic subset of Z (see Exercise 8.9). (a) Show that every periodic point of T is almost-periodic. (b) Show that if X is compact and (X, T ) is minimal then every point x ∈ X is almost-periodic.

Exercises

155

(c) Suppose that X is a compact Hausdorff space and x ∈ X is almost-periodic. Let Y denote the closure in X of the orbit of x. Show that the dynamical system (Y, T ) is minimal. 8.11 Let n, m ∈ Z. Show that the shift ((Rn )Z , σ) embeds in the shift ((Rm )Z , σ) if and only if n ≤ m. 8.12 Let X be a compact metrizable space and T : X → X a homeomorphism. Let d be a metric on X compatible with the topology. The dynamical system (X, T ) is called distal if the following condition is satisfied: given any pair of distinct points x and y in X , there exists a real number ε > 0 such that d(T n (x), T n (y)) ≥ ε for all n ∈ Z. (a) Show that this definition does not depend on the choice of the metric d. (b) Show that if the dynamical system (X, T ) is both distal and minimal, then it embeds in the shift (RZ , σ). 8.13 Let S1 := {(x1 , x2 ) ∈ R2 | x12 + x22 = 1} denote the unit circle in R2 . Let T : S1 → S1 be the half-turn given by T (x1 , x2 ) = −(x1 , x2 ). Show that the dynamical system (S1 , T ) does not embed in the shift (RZ , σ). Hint: observe that the dynamical system (Per 2 (T ), T ) does not embed in the dynamical system (Per 2 (σ), σ). 8.14 (cf. [51, Example 4.1]). Let Y be a compact metrizable space and n ≥ 2 an integer. Consider the product space X := Y × {0, 1, . . . , n − 1} and the homeomorphism T : X → X defined, for all x = (y, k) ∈ X , by T (x) :=

(y, k + 1) (y, 0)

if k ≤ n − 2 if k = n − 1.

Show that the dynamical system (X, T ) embeds in the shift (RZ , σ) if and only if Y is topologically embeddable in Rn . 8.15 Let X be a compact metrizable space with dim(X ) = n < ∞ and T : X → X a homeomorphism. Suppose that every orbit of T contains at least 6n +1 distinct points. Show that the dynamical system (X, T ) embeds in the shift (RZ , σ). Hint: observe that Lemma 8.3.3 remains valid for m = 2n + 1.

Chapter 9

Amenable Groups

This chapter is devoted to the class of amenable groups, a class of groups which contains all finite groups and all abelian groups and which is closed under several group operations, in particular taking subgroups, taking extensions, and taking direct limits. Countable amenable groups can be characterized by the existence of Følner sequences, i.e., sequences of non-empty finite subsets of the group that are asymptotically invariant under translations. Using such Følner sequences, it is possible to define the average value of an invariant subadditive function on the set of finite subsets of the groups (see Theorem 9.4.1). This will be used in the next chapter for extending the definition of mean topological dimension to continuous actions of countable amenable groups.

9.1 Følner Sequences From now on, we prefer to use the notation |A| instead of # A to denote the cardinality of a finite set A. Let G be a group. We assume that the group operation on G is denoted multiplicatively (additive notation will be used in some examples, e.g., for G = Zd ). The identity element of G is denoted 1G . We recall that the subgroup generated by a subset A ⊂ G is the smallest subgroup A ⊂ G containing A. It consists of all elements g ∈ G that can be written in the form g = a1ε1 · · · anεn where n ≥ 0, ai ∈ A and εi ∈ {−1, 1} for all 1 ≤ i ≤ n. One says that the group G is finitely generated if there exists a finite subset A ⊂ G such that A = G. Note that every finitely generated group is countable. If A is a subset of G and g ∈ G, the left-translate of A by g is the set g A := {ga | a ∈ A} ⊂ G. Similarly, the right-translate of A by g is the set Ag := {ag | a ∈ A} ⊂ G. Observe that A\g A consists of all elements of A that are moved out of A under left-translation by g −1 . When A is finite, the sets A and g A have the same cardinality © Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_9

157

158

9 Amenable Groups

so that |A\g A| = |g A\A| = |A| − |A ∩ g A|. If A is a non-empty finite subset of G, the ratio |A\g A| |g A\A| |A ∩ g A| = =1− |A| |A| |A| is the proportion of elements in A that are moved out of A by left-translation by g −1 . So, heuristically, this proportion measures the lack of invariance of A when it is left-translated by g −1 . A Følner sequence for a group G is a sequence of non-empty finite subsets of G that are “asymptotically left-invariant”. The precise definition goes as follows. Definition 9.1.1 Let G be a group. One says that a sequence (Fn )n≥1 of non-empty finite subsets of G is a Følner sequence for G if one has lim

n→∞

|Fn \g Fn | =0 |Fn |

for all g ∈ G. Proposition 9.1.2 If a group G admits a Følner sequence, then G is countable. Proof Suppose that (Fn )n≥1 is a Følner sequence for the group G. Consider the finite subsets An of G defined by An := {x y −1 : x, y ∈ Fn }. This implies that Given g ∈ G, we have |Fn \g Fn | ≤ |Fn |/2 for n large enough. g Fn meets Fn and hence that g ∈ An . It follows that G = n≥1 An . Therefore G is countable. Definition 9.1.3 One says that a countable group is amenable if it admits a Følner sequence. Proposition 9.1.4 Every finite group is amenable. Proof If G is a finite group, then the constant sequence Fn = G is a Følner sequence for G since Fn \g Fn = G\G = ∅ for all g ∈ G. An example of an infinite amenable group is provided by the additive group Z of integers. Proposition 9.1.5 The group Z is amenable.

9.1 Følner Sequences

159

Proof Consider the sequence (Fn )n≥1 defined by Fn := {0, 1, . . . , n − 1} ⊂ Z. Let us fix g ∈ Z. Then, for all n ≥ |g|, we have ⎧ ⎪ if g ≥ 1, ⎨{0, 1, . . . , g − 1} Fn \(g + Fn ) = ∅ if g = 0, ⎪ ⎩ {g + n, g + n + 1, . . . , n − 1} if g ≤ −1. and hence

|Fn \(g + Fn )| |g| = , |Fn | n

which converges to 0 as n goes to infinity. We deduce that the sequence (Fn )n≥1 is a Følner sequence for Z and hence that Z is amenable. Proposition 9.1.6 Suppose that G 1 and G 2 are countable amenable groups. Then the group G = G 1 × G 2 is also amenable. Proof Let (An )n≥1 be a Følner sequence for G 1 and let (Bn )n≥1 be a Følner sequence for G 2 . Consider the non-empty subsets Fn ⊂ G defined by Fn := An × Bn . For every g = (g1 , g2 ) ∈ G, we have Fn \g Fn = ((An \g1 An ) × Bn ) ∪ (An × (Bn \g2 Bn )) and hence |Fn \g Fn | = | ((An \g1 An ) × Bn ) ∪ (An × (Bn \g2 Bn )) | ≤ |(An \g1 An ) × Bn | + |An × (Bn \g2 Bn )| = |An \g1 An ||Bn | + |An ||Bn \g2 Bn |. Dividing by |Fn | = |An ||Bn |, we obtain |An \g1 An | |Bn \g2 Bn | |Fn \g Fn | ≤ + . |Fn | |An | |Bn | The right-hand side of the preceding inequality tends to 0 as n goes to infinity since (An ) and (Bn ) are Følner sequences for G 1 and G 2 respectively. It follows that lim

n→∞

|Fn \g Fn | = 0. |Fn |

This shows that the sequence (Fn )n≥1 is a Følner sequence for G and hence that G is amenable.

160

9 Amenable Groups

Remark 9.1.7 Note that the preceding proof gives us an explicit way for constructing a Følner sequence for G 1 × G 2 if we are given Følner sequences for G 1 and G 2 . Corollary 9.1.8 The group Zd is amenable for every integer d ≥ 0. Proof This immediately follows from Propositions 9.1.5 and 9.1.6 by induction on d. Corollary 9.1.9 Every finitely generated abelian group is amenable. Proof This follows from Proposition 9.1.6, Corollary 9.1.8, and Proposition 9.1.4 since it is known that if G is a finitely generated abelian group then there exist an integer d ≥ 0 and a finite group T such that G is isomorphic to Zd × T . Remark 9.1.10 If d ≥ 1 and T is a finite group, an explicit Følner sequence for Zd × T is provided by the sequence (Fn )n≥1 , where Fn := {0, 1, . . . , n − 1}d × T.

9.2 Amenable Groups In the previous section, the notion of amenability has been only defined for countable groups via Følner sequences. Although we are here mainly interested in countable groups, it is very convenient to consider the class of all amenable groups, countable or not. In order to extend the definition of amenability to uncountable groups, we shall use the following characterization of countable amenable groups. Lemma 9.2.1 Let G be a countable group. Then the following conditions are equivalent: (a) G admits a Følner sequence; (b) for every finite subset S ⊂ G and every ε > 0, there exists a non-empty finite subset F ⊂ G such that |F\s F| ≤ ε|F| for all s ∈ S. Proof Suppose that (Fn )n≥1 is a Følner sequence for G. Let S be a finite subset of G and ε > 0. For each s ∈ S, we can find an integer n(s) such that |Fn \s Fn | ≤ ε|Fn | for all n ≥ n(s). Then F := Fm , where m := maxs∈S n(s), satisfies |F\s F| ≤ ε|F| for all s ∈ S. This shows that (a) implies (b). Conversely, suppose that (b) is satisfied. As G is countable, we can write G = {g1 , g2 , g3 , . . . }. Choose some sequence of positive real numbers (εn )n≥1 that converges to 0. By applying (b) with S = {g1 , g2 , . . . , gn } and ε = εn , we deduce that there exists a finite non-empty subset Fn ⊂ G such that |Fn \gk Fn |/|Fn | ≤ εn for all 1 ≤ k ≤ n. Fixing k, it follows that |Fn \gk Fn |/|Fn | tends to 0 as n goes to infinity. Therefore the sequence (Fn )n≥1 is a Følner sequence for G. This shows that (b) implies (a).

9.2 Amenable Groups

161

Definition 9.2.2 One says that a (possibly uncountable) group G is amenable if it satisfies the following condition: for every finite subset S ⊂ G and every ε > 0, there exists a non-empty finite subset F ⊂ G such that |F\s F| ≤ ε|F| for all s ∈ S. In the case when G is countable, it follows from Lemma 9.2.1 that the preceding definition is equivalent to the one given in Sect. 9.1, namely the existence of a Følner sequence for G. Recall however that uncountable groups do not admit Følner sequence by Proposition 9.1.2. The condition in Definition 9.2.2 that characterizes amenable groups is known as the Følner condition. Proposition 9.2.3 Let G be a group. Suppose that there exists a family (Hi )i∈I of amenable subgroups of G satisfying the following conditions: (C1) G = i∈I Hi ; (C2) for all i 1 , i 2 ∈ I , there exists j ∈ I such that Hi1 ∪ Hi2 ⊂ H j . Then G is amenable. Proof Let S be a finite subset of G and ε > 0. By (C1), we can find, for each s ∈ S, an index i(s) ∈ I such that s ∈ Hi(s) . Now we deduce from (C2) that there exists j ∈ I such that s∈S Hi(s) ⊂ H j . As H j is amenable and S ⊂ H j , there exists a non-empty finite subset F ⊂ H j ⊂ G such that |F\g F| ≤ ε|F| for all g ∈ S. This shows that G is amenable. Corollary 9.2.4 Let G be a group. Suppose that there exists a non-decreasing sequence H0 ⊂ H1 ⊂ H2 ⊂ . . . of amenable subgroups of G such that G =

n≥0

Hn . Then G is amenable.

Definition 9.2.5 Let P be a property of groups. A group G is said to be locally P if every finitely generated subgroup of G has property P. Corollary 9.2.6 Let G be a group. Suppose that every finitely generated subgroup of G is amenable. Then G is amenable. In other words, every locally amenable group is amenable. Proof Denote by S the set consisting of all finitely generated subgroups of G. Then the family (H ) H ∈S clearly satisfies the conditions of Proposition 9.2.3. Corollary 9.2.7 Every abelian group is amenable. Proof This is an immediate consequence of Corollary 9.2.6 since we already know that every finitely generated abelian group is amenable by Corollary 9.1.9. Example 9.2.8 The additive group R of real numbers is amenable since it is abelian. This provides an example of an uncountable amenable group.

162

9 Amenable Groups

A group G is called a torsion group if every element of G has finite order. Every locally finite group is a torsion group. Indeed, if a group G is locally finite then the group generated by any element x ∈ G must be finite. Conversely, every abelian torsion group is locally finite. This follows from the above mentioned result about the classification of finitely generated abelian groups, namely that every finitely generated abelian group is isomorphic to Zd × T for some integer d ≥ 0 and some finite group T , since it immediately implies that every finitely generated abelian torsion group is finite. Example 9.2.9 The group Q/Z (which is isomorphic to the multiplicative group formed by all the roots of unity in C) is locally finite. Note that Q/Z is countably infinite. Another example of a countably infinite, locally finite, abelian group is provided by the additive group K [x] of polynomials over a finite field K . Example 9.2.10 Let X be a set. Consider the symmetric group of X , that is, the group Sym(X ) consisting of all permutations σ : X → X with the composition of maps as the group operation. The support of an element σ ∈ Sym(X ) is the subset supp(σ) ⊂ X consisting of all x ∈ X such that σ(x) = x. Observe that supp(σ −1 ) = supp(σ) and that supp(σ1 σ2 ) ⊂ supp(σ1 ) ∪ supp(σ2 ) for all σ, σ1 , σ2 ∈ Sym(X ). Let Sym0 (X ) ⊂ Sym(x) denote the set of permutations of X whose support is finite. Then Sym0 (X ) is a locally finite subgroup of Sym(X ). It is non-abelian as soon as X contains more than two elements. Observe that Sym0 (X ) is countably infinite and isomorphic to Sym(N) whenever X is countably infinite. Proposition 9.2.11 Every locally finite group is amenable. Proof This immediately follows from Proposition 9.1.4 and Corollary 9.2.6.

Given two subsets A and B of a group G, the subset AB ⊂ G is defined by AB := {ab | a ∈ A, b ∈ B} =

aB =

a∈A

Ab.

b∈B

The following characterization of amenability is sometimes easier to handle than the original Følner condition. Lemma 9.2.12 Let G be a group. Then the following conditions are equivalent: (a) G is amenable; (b) for every finite subset S ⊂ G and every ε > 0, there exists a non-empty finite subset F ⊂ G such that |S F\F| ≤ ε|F|. Proof Suppose first that G is amenable. Let S be a finite subset of G and ε > 0. Since G is amenable, there exists a non-empty finite subset F ⊂ G such that |F\s F| ≤

ε |F| for all s ∈ S. |S|

9.2 Amenable Groups

163

It follows that |S F\F| = (s F\F) s∈S ≤ |s F\F| s∈S

=

|F\s F|

s∈S

≤

ε |S| s∈S

= ε. This shows that (a) implies (b). Conversely, suppose (b). Let S be a finite subset of G and ε > 0. Condition (b) implies the existence of a non-empty finite subset F ⊂ G such that |S F\F| ≤ ε|F|. Since, for every s ∈ S, we have that s F\F ⊂ S F\F, it follows that |F\s F| = |s F\F| ≤ |S F\F| ≤ ε|F|.

This shows that G is amenable.

We shall use the preceding lemma for proving that the class of amenable groups is closed under taking subgroups. Proposition 9.2.13 Every subgroup of an amenable group is amenable. Proof Let H be a subgroup of an amenable group G. Let S be a finite subset of H and ε > 0. As G is amenable, it follows from Lemma 9.2.12 that there exists a non-empty finite subset E ⊂ G such that |S E\E| ≤ ε|E|.

(9.2.1)

Let T ⊂ G be a complete set of representatives of the right cosets of H in G. This means that every g ∈ G can be uniquely written in the form g = ht with h ∈ H and t ∈ T . For each t ∈ T , denote by E t the subset of H consisting of all h ∈ H such

that ht ∈ E. Consider

also the subset T ⊂ T defined by T := {t ∈ T | E t = ∅}. We then have E = t∈T E t t and hence |E| =

|E t t| =

t∈T

Since S ⊂ H , we have that S E =

|E t |.

t∈T

t∈T

S E\E =

S E t t and hence

t∈T

(S E t \E t )t.

(9.2.2)

164

9 Amenable Groups

This gives us |S E\E| =

|(S E t \E t )t| =

t∈T

|S E t \E t |.

t∈T

Using (9.2.1) and (9.2.2), we then get

|S E t \E t | ≤ ε

t∈T

|E t |.

t∈T

This last inequality implies that there exists t0 ∈ T such that |S E t0 \E t0 | ≤ ε|E t0 |. As E t0 is a non-empty finite subset of H , we deduce from Lemma 9.2.12 that H is amenable. One says that a group G is an extension of a group A by a group B if there exists a normal subgroup H of G such that H is isomorphic to A and G/H is isomorphic to B. Our next result is that the class of amenable groups is closed under extensions. Proposition 9.2.14 Let G be a group. Suppose that there exists a normal subgroup H of G such that the group H and the quotient group K = G/H are both amenable. Then G is amenable. Proof Let S be a finite subset of G and ε > 0. We want to show that there exists a non-empty finite subset F ⊂ G such that |F\s F| ≤ ε|F| for all s ∈ S. Denote by q : G → K the quotient homomorphism, i.e., the map defined by q(g) = g H = H g for all g ∈ G. Let T = q(S). As K is amenable, there exists a non-empty finite subset C ⊂ K such that |C\tC| ≤

ε |C| for all t ∈ T. 2

(9.2.3)

Let us choose a set of representatives for the cosets of H that are in C, i.e., a set A ⊂ G such that q induces a bijection from A onto C. Now let R denote the set consisting of all h ∈ H for which there exist elements a, a ∈ A and s ∈ S such that h = a −1 sa , that is, R := A−1 S A ∩ H . Observe that R is a finite subset of H (of cardinality bounded above by |A|2 |S|). As H is amenable, there exists a non-empty finite subset B ⊂ H such that |B\h B| ≤

ε |B| for all h ∈ R. 2|R|

(9.2.4)

Consider now the subset F ⊂ G defined by F := AB. Note that q(F) = C. Observe also that each element g ∈ F can be uniquely written in the form x = ab with a ∈ A and b ∈ B. Consequently, we have that

9.2 Amenable Groups

165

|F| = |A||B|.

(9.2.5)

Let us fix some element s ∈ S and let t := q(s) ∈ T . In order to bound from above the cardinality of F\s F, we observe that F\s F is the disjoint union of the sets E 1 and E 2 defined by / tC} E 1 := {g ∈ F\s F | q(g) ∈ and E 2 := {g ∈ F\s F | q(g) ∈ tC}. If g ∈ E 1 and we write g = ab, with a ∈ A and b ∈ B, then q(a) = q(g) ∈ C\tC. Since q is injective on A, we deduce that |E 1 | ≤ |C\tC||B| ε ≤ |C||B| 2 ε = |A||B| 2 ε = |F| 2

(by (9.2.3))

(by (9.2.5)).

Suppose now that g ∈ E 2 and write again g = ab with a ∈ A and b ∈ B. We then have q(a) = q(g) ∈ tC = q(s A). It follows that we can find h ∈ H and a ∈ A such that g = ab = sa h −1 b. ah = sa . Observe that h ∈ R by definition of R. We then get This last equality implies b ∈ / h B since g ∈ / s F. Thus b ∈ h∈R (B\h B). This gives us |E 2 | ≤ |A| (B\h B) h∈R

≤ |A| |B\h B| h∈R

≤ |A||R|

ε |B| 2|R|

ε |A||B| 2 ε = |F| 2

(by (9.2.4))

=

(by (9.2.5)).

Combining the above results, we finally get |F\s F| = |E 1 | + |E 2 | ≤

ε ε |F| + |F| = ε|F|. 2 2

166

9 Amenable Groups

This shows that F has the required properties. Thus G is amenable.

One says that a group G is a semidirect product of a group G 1 with a group G 2 if G contains a normal subgroup H isomorphic to G 1 and a subgroup K isomorphic to G 2 such that G = K H and K ∩ H = {1G }. As the quotient group G/H is then isomorphic to K and hence to G 2 , an immediate consequence of Proposition 9.2.14 is the following: Corollary 9.2.15 If a group G is a semidirect product of two amenable groups then G is amenable. A group G is called metabelian if it is an extension of an abelian group by an abelian group, i.e., G contains a normal subgroup H such that both H and G/H are abelian. From Corollary 9.2.7 and Proposition 9.2.14, we get: Corollary 9.2.16 Every metabelian group is amenable.

Example 9.2.17 Let K be a field. Consider the group G consisting of all affine transformations of K , i.e., all transformations of the form x → ax + b (x ∈ K ), where a ∈ K = K \{0} and b ∈ K , with the composition of maps as the group operation. The translations x → x + b, b ∈ K , form an abelian normal subgroup T of G isomorphic to the additive group K and the quotient group G/T is isomorphic to the multiplicative group K = K \{0} (the map that sends the affine transformation x → ax +b to a is a surjective group homomorphism from G onto the multiplicative group K with kernel T ). Therefore G is metabelian. Note that G is also the semidirect product of its normal subgroup T with the subgroup of homotheties x → ax (a ∈ K ). Example 9.2.18 Let G denote the group of affine transformations of the real line R and let T denote the normal subgroup of G consisting of all translations x → x + b (see Example 9.2.17). Fix an integer n ≥ 2 and consider the subgroup G n of G generated by the translation t : x → x + 1 and the homothety h : x → nx. The group G n consists of all affine transformations of the form x → ax + b, where a = n k , k ∈ Z, is an integral power of n and b ∈ Z[1/n] is an n-adic rational (i.e., a rational of the form mn k , where m, k ∈ Z). The translation subgroup Tn = G n ∩ T is isomorphic to the additive group Z[1/n]. It is normal in G n with quotient group G n /Tn infinite cyclic. Therefore G n is a countably infinite metabelian group. Note that G n is not abelian since ht = th. The group G n belongs to the family of BaumslagSolitar groups and is denoted B S(1, n). Example 9.2.19 Let R be a commutative ring. The Heisenberg group is the subgroup H R of GL3 (R) consisting of all matrices of the form M(a, b, c), where ⎛

⎞ 1ac M(a, b, c) := ⎝0 1 b⎠ (a, b, c ∈ R). 001

9.2 Amenable Groups

167

The map defined by M(a, b, c) → (a, b) is a surjective group homomorphism from H R onto the additive group R 2 whose kernel is isomorphic to the additive group R. Thus H R is a metabelian group. When R = {0}, it is not abelian since the matrices M(1, 0, 0) and M(0, 1, 0) do not commute. The integral Heisenberg group HZ provides another example of a countably infinite non-abelian metabelian group. Let G be a group. The commutator of two elements g and h of G is the element [g, h] ∈ G defined by [g, h] := ghg −1 h −1 . The subgroup of G generated by all commutators [g, h], with g, h ∈ G, is called the derived subgroup of G and is denoted D(G). If α is a group automorphism of G, then one clearly has α([g, h]) = [α(g), α(h)] for all g, h ∈ G. This implies α(D(G)) = D(G), that is, D(G) is invariant by α. In particular D(G) is invariant by every inner automorphism of G, i.e., D(G) is normal in G. Note that the quotient group G/D(G) is abelian since gh = [g, h]hg for all g, h ∈ G. Moreover, if H is a normal subgroup of G such that G/H is abelian then D(G) ⊂ H . The derived series of G is the sequence (D n (G))n≥0 of subgroups of G inductively defined by D 0 (G) := G and D n+1 (G) := D(D n (G)) for all n ≥ 0. One has G = D 0 (G) ⊃ D(G) = D 1 (G) ⊃ D 2 (G) ⊃ . . . with D n+1 (G) normal in D n (G) and D n (G)/D n+1 (G) abelian for all n ≥ 0. The group G is said to be solvable if there is an integer n ≥ 0 such that D n (G) = {1G }. The smallest integer n ≥ 0 such that D n (G) = {1G } is then called the solvability degree of G. Note that the solvable groups of solvability degree 0 (resp. 1, resp. 2) are the trivial groups (resp. the non-trivial abelian groups, resp. the non-abelian metabelian groups). Note also that if a group G is solvable of solvability degree n then any subgroup of G and any quotient group of G is solvable with solvability degree ≤ n. Example 9.2.20 Let K be a field and let G denote the subgroup of GLn (K ) consisting of all upper triangular matrices, i.e., all matrices of the form M = (m i j )1≤i, j≤n with m i j ∈ K for all 1 ≤ i, j ≤ n, m i j = 0 for i > j, and m ii = 0 for 1 ≤ i ≤ n. Then G is a solvable group. To see this, consider, for each integer k ≥ 0, the subset E k ⊂ GLn (K ) consisting of all matrices M = (m i j ) such that m i j = δi j for all i > j − k, where δi j is the Kronecker symbol defined by δi j = 1 if i = j and δi j = 0 if i = j. In other words, E k consists of all matrices in GLn (K ) whose entries located below the diagonal line i = j − k coincide with the corresponding entries of the identity matrix In . Note that E 0 = G. An easy computation shows that if M, N ∈ E k then [M, N ] ∈ E k+1 . By induction on k, this implies D k (G) ⊂ E k for all k ≥ 0. In particular, we have D n (G) ⊂ E n . As E n = {In }, we deduce that G is solvable. Theorem 9.2.21 Every solvable group is amenable.

168

9 Amenable Groups

Proof If G is solvable of solvability degree 0 then G is trivial and hence amenable. Suppose now that G is solvable of solvability degree n ≥ 1. Observe that D(G) G. From the definition of the derived series, we see that D(G) is solvable of solvability degree ≤ n − 1. As G/D(G) is abelian and hence amenable by Corollary 9.2.7, we conclude that G is amenable by applying Proposition 9.2.14 and induction on n. If P is a property of groups, one says that a group G is virtually P if G contains a subgroup of finite index that satisfies P. For example, a virtually abelian group is a group that contains an abelian subgroup of finite index. Corollary 9.2.22 Every virtually amenable group is amenable. For the proof, we need the following classical result from group theory. Lemma 9.2.23 Let G be a group. Suppose that H is a subgroup of finite index of G. Then there exists a normal subgroup of finite index K of G such that K ⊂ H . Proof The group G acts by left translation on the set G/H of left cosets of H (the left translate by g ∈ G of the coset x H , x ∈ G, is the coset gx H ). This action is described by a homomorphism ρ : G → Sym(G/H ), where Sym(G/H ) is the group of permutations of G/H . The kernel K of ρ is of finite index in G since G/H is finite (the index of K in G divides | Sym(G/H )| = n!, where n = [G : H ] = |G/H |). We have K ⊂ H since if g ∈ K then g fixes each coset and in particular H . Proof of Corollary 9.2.22 Let G be a virtually amenable group. This means that G contains an amenable subgroup H of finite index. By Lemma 9.2.23, there exists a normal subgroup of finite index K of G such that K ⊂ H . The group K is amenable since every subgroup of an amenable group is itself amenable by Proposition 9.2.13. As G/K is finite and hence amenable by Proposition 9.1.4, we conclude that G is amenable by applying Proposition 9.2.14. Corollary 9.2.24 Every locally virtually amenable group is amenable. Proof This follows from Corollaries 9.2.6 and 9.2.22.

Corollary 9.2.25 Every locally virtually solvable group is amenable. Proof This immediately follows from Theorem 9.2.21 and Corollary 9.2.24.

9.3 Examples of Non-amenable Groups The goal of this section is to provide examples of groups that are not amenable. The reader is assumed to have some familiarity with free groups. Let us start by recalling some basic facts about them.

9.3 Examples of Non-amenable Groups

169

One says that a group G is free if there exists a subset X ⊂ G such that the pair (G, X ) satisfies the following universal property: given any group H and any map f : X → H , there exists a unique group homomorphism ϕ : G → H such that ϕ(x) = f (x) for all x ∈ X . One then says that X is a base of the free group G. It can be shown that if G is a group and X ⊂ G, then G is free with base X if and only if every g ∈ G can be uniquely written in the form g = a1 a2 . . . an ,

(9.3.1)

where n ≥ 0, ai ∈ X ∪ X −1 for all 1 ≤ i ≤ n, and ai+1 = ai−1 for all 1 ≤ i ≤ n − 1. This is known as the normal form of the element g. Given a set X , there always exists a group with base X and such a group is unique up to a unique isomorphism fixing X pointwise. This is the reason why it is a common abuse to speak of the free group with base X to designate any of the free groups with base X . Proposition 9.3.1 If X is a set with more than one element, then the free group with base X is non-amenable. Proof Let X be a set with more than one element and let G denote the free group with base X . Let x and y be two distinct elements in X . Consider the set S ⊂ G defined by S := {x, y, x −1 , y −1 }. If G were amenable, then for every ε > 0 we could find a non-empty finite subset F ⊂ G satisfying |F\s F| ≤ ε|F| for all s ∈ S. (9.3.2) For each s ∈ S, let G s denote the subset of G consisting of all elements g = 1G whose normal form (9.3.1) starts with a1 = s −1 . The sets G s , s ∈ S, are pairwise disjoint. This implies in particular that

|F ∩ G s | ≤ |F|.

(9.3.3)

s∈S

On the other hand, for each s ∈ S, we have that |F| = |F\G s | + |F ∩ G s | = |s(F\G s )| + |F ∩ G s |. Now observe that s(G\G s ) ⊂ G s −1 so that s(F\G s ) ⊂ (s F\F) ∪ (F ∩ G s −1 )

(9.3.4)

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9 Amenable Groups

and hence |s(F\G s )| ≤ |s F\F| + |F ∩ G s −1 | = |F\s F| + |F ∩ G s −1 | ≤ ε|F| + |F ∩ G s −1 |

(by (9.3.2)).

By using (9.3.4), we deduce that |F| ≤ ε|F| + |F ∩ G s −1 | + |F ∩ G s | for all s ∈ S. After summing up over all s ∈ S, this gives us 4|F| ≤ 4ε|F| +

|F ∩ G s −1 | + |F ∩ G s |

s∈S

= 4ε|F| + 2

|F ∩ G s |.

s∈S

Combining with (9.3.3), we finally get 4|F| ≤ 4ε|F| + 2|F| and hence |F| ≤ 2ε|F|, which yields a contradiction for ε < 1/2. This shows that G is not amenable. Remark 9.3.2 Suppose that G is a free group with base X . As every abelian group is amenable by Corollary 9.2.7, we deduce from Proposition 9.3.1 that G is non-abelian if X has more than one element. In fact, this can be shown directly by observing that if X contains two distinct elements x and y then x y = yx by uniqueness of normal forms in G. Note that if X is empty (resp. reduced to one single element) then G is trivial (resp. infinite cyclic). Thus, the following conditions are all equivalent: (1) G is non-amenable; (2) G is non-abelian; (3) X contains more than one element. Combining Propositions 9.2.13, 9.3.1, and the above remark we get the following result. Corollary 9.3.3 If a group G contains a non-abelian free subgroup then G is nonamenable. Corollary 9.3.3 may be used for showing that certain matrix groups are not amenable. Here is an example. Recall that the group SLd (Z) is the multiplicative group of d × d matrices with entries in Z and determinant 1. Corollary 9.3.4 The group SLd (Z) is non-amenable for d ≥ 2. Proof As the group SL2 (Z) embeds into SLd (Z) for any d ≥ 2, it suffices to show that SL2 (Z) is non-amenable.

9.3 Examples of Non-amenable Groups

171

Consider the matrices P, Q ∈ SL2 (Z) defined by

13 10 P := and Q := . 01 31 Let G be the free group based on a set X = {x, y}, with x = y. By the universal property of free groups, there is a unique group homomorphism ϕ : G → SL2 (Z) satisfying ϕ(x) = P and ϕ(y) = Q. Let us show that ϕ is injective. Suppose not. Then there exists an element g = 1G in G such that ϕ(g) = I , where I ∈ SL2 (Z) denotes the identity matrix. Let us write g in normal form: g = a1 a2 . . . an where n ≥ 1, ai ∈ {x, y, x −1 , y −1 } for all 1 ≤ i ≤ n, and ai+1 = ai−1 for all 1 ≤ i ≤ n − 1. After grouping equal consecutive factors, this can be written as g = u n1 1 u n2 2 · · · u nk k where k ≥ 1, u i ∈ {x, y} and n i ∈ Z\{0} for all 1 ≤ i ≤ k, and u i = u i+1 for all 1 ≤ i ≤ k − 1. Setting Ui = ϕ(u i ) for 1 ≤ i ≤ k, we obtain I = U1n 1 U2n 2 · · · Ukn k ,

(9.3.5)

where k ≥ 1, Ui ∈ {P, Q} and n i ∈ Z\{0} for all 1 ≤ i ≤ k, and Ui = Ui+1 for all 1 ≤ i ≤ k − 1. 2 To prove that this is impossible, we use the natural action of SL 2 (Z)on Z , that ab is, the action given by gv = (av1 + bv2 , cv1 + dv2 ) for all g = ∈ SL2 (Z) cd and v = (v1 , v2 ) ∈ Z2 . Indeed, consider the subsets T1 , T2 ⊂ Z2 defined by T1 := {(v1 , v2 ) ∈ Z2 | |v1 | > |v2 |} and T2 := {(v1 , v2 ) ∈ Z2 | |v2 | > |v1 |} and the point b := (1, 1) ∈ Z2 (see Fig. 9.1). We claim that, for all n ∈ Z\{0}, the following hold: (P1) (P2) (P3) (P4)

P n b ∈ T1 ; Q n b ∈ T2 ; P n v ∈ T1 for all v ∈ T2 ; Q n v ∈ T2 for all v ∈ T1 .

Indeed, let n ∈ Z\{0}. Property (P1) is satisfied since P n b = (1 + 3n, 1) and |1 + 3n| ≥ 2. On the other hand, if v = (v1 , v2 ) ∈ T2 , then we have P n v = (v1 + 3nv2 , v2 ) and, by the triangle inequality, |v1 + 3nv2 | ≥ 3|n||v2 | − |v1 | ≥ 3|v2 | − |v1 | > 2|v2 |,

172

9 Amenable Groups

Fig. 9.1 The ping-pong table

so that P n v ∈ T1 . This shows (P3). Properties (P2) and (P4) are obtained similarly by exchanging coordinates. This establishes our claim. Now, we distinguish two cases. Suppose first that Uk = P. Then, by using (P1), (P3), and (P4), we successively get n

n

n

n

n

n

k−1 k−2 k−1 k−3 k−2 k−1 Ukn k b ∈ T1 , Uk−1 Ukn k b ∈ T2 , Uk−2 Uk−1 Ukn k b ∈ T1 , Uk−3 Uk−2 Uk−1 Ukn k b ∈ T2 , . . . .

On the other hand, if Uk = Q, then we deduce from (P2), (P3), and (P4) that n

n

n

n

n

n

k−1 k−2 k−1 k−3 k−2 k−1 Ukn k b ∈ T2 , Uk−1 Ukn k b ∈ T1 , Uk−2 Uk−1 Ukn k b ∈ T2 , Uk−3 Uk−2 Uk−1 Ukn k b ∈ T1 , . . . .

Thus, in both cases, we have that U1n 1 U2n 2 . . . Ukn k b ∈ T1 ∪ T2 . As b ∈ / T1 ∪ T2 , this gives us U1n 1 U2n 2 . . . Ukn k b = b, which contradicts (9.3.5). Consequently, the group homomorphism ϕ : G → SL2 (Z) is injective. This implies that ϕ(G) is a non-abelian free subgroup of SL2 (Z). By applying Corollary 9.3.3, we deduce that SL2 (Z) is non-amenable. Remark 9.3.5 The key step in the above proof is an example of application of the ping-pong principle. It may be visualized by thinking of T1 and T2 as the two halves of a tennis table and observing that in the sequence n

n

n

n

n

n

k−1 k−2 k−1 k−3 k−2 k−1 Ukn k b, Uk−2 Uk−1 Ukn k b, Uk−3 Uk−2 Uk−1 Ukn k b, . . . , Ukn k b, Uk−1

the points lie alternatively in each of these halves so that they cannot return to the initial position b because b is outside of the table.

9.4 The Subadditive Convergence Theorem for Amenable Groups

173

9.4 The Subadditive Convergence Theorem for Amenable Groups The goal of this section is to establish the following result, which is an analogue of Proposition 6.2.3 for countable amenable groups. Theorem 9.4.1 (Ornstein-Weiss lemma) Let G be a countable amenable group and let F = (Fn )n≥1 be a Følner sequence for G. Let P f in (G) denote the set of all finite subsets of G. Suppose that h : P f in (G) → R is a real-valued map satisfying the following conditions: (H1) h is subadditive, i.e., one has h(A ∪ B) ≤ h(A) + h(B) for all A, B ∈ P f in (G); (H2) h is right-invariant, i.e., one has h(Ag) = h(A) for all g ∈ G and A ∈ P f in (G). Then the limit λ = lim

n→∞

h(Fn ) |Fn |

exists and one has 0 ≤ λ < ∞. Moreover, the limit λ does not depend on the choice of the Følner sequence F for G. The proof of Theorem 9.4.1 is rather long and technical. It is based on several auxiliary results. Lemma 9.4.2 Let A and B be finite subsets of a group G. Then one has

|Ag ∩ B| = |A||B|.

(9.4.1)

g∈G

Proof For E ⊂ G, denote by χ E : G → R the characteristic map of E, i.e., the map defined by χ E (g) = 1 if g ∈ E and χ E (g) = 0 otherwise. Then we have

|Ag ∩ B| =

g∈G

g∈G

=

g∈G

=

g ∈G

⎛ ⎝

⎞ χ Ag∩B (g )⎠

g ∈G

⎛ ⎝

g ∈G

⎛ ⎝

g∈G

⎞ χ Ag (g )χ B (g )⎠ ⎞ χ Ag (g )⎠ χ B (g )

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9 Amenable Groups

=

⎛

⎝

g ∈G

=

⎞ χ A (g g −1 )⎠ χ B (g )

g∈G

|A|χ B (g )

g ∈G

= |A|

χ B (g )

g ∈G

= |A||B|.

This shows (9.4.1).

Definition 9.4.3 Let X be a set and ε > 0. A family (Ai )i∈I of finite subsets of X is called ε-disjoint if there exists a family (Bi )i∈I of pairwise disjoint subsets of X such that Bi ⊂ Ai and |Ai \Bi | ≤ ε|Ai | for all i ∈ I . Lemma 9.4.4 Let X be a set and ε > 0. Suppose that (Ai )i∈I is an ε-disjoint finite family of finite subsets of X . Then one has (1 − ε)

i∈I

|Ai | ≤ Ai .

(9.4.2)

i∈I

Proof Let (Bi )i∈I be a family of subset of X as in Definition 9.4.3. We then have (1 − ε)

i∈I

|Ai | ≤

i∈I

|Bi | = Bi ≤ Ai . i∈I

i∈I

This shows (9.4.2).

Definition 9.4.5 Let and K be subsets of a group G. The K -interior of is the subset Int K () ⊂ G defined by Int K () := {g ∈ G | K g ⊂ }. The K -closure of is the subset Cl K () ⊂ G defined by Cl K () := {g ∈ G | K g ∩ = ∅}. The K -boundary of is the subset ∂ K () ⊂ G defined by ∂ K () := Cl K ()\ Int K ().

9.4 The Subadditive Convergence Theorem for Amenable Groups

175

In other words, Int K () (resp. Cl K (), resp. ∂ K ()) is the set consisting of all g ∈ G such that the right-translate of K by g is contained in (resp. meets , resp. meets both and G\). Lemma 9.4.6 Let G be a group. Then one has (i) (ii) (iii) (iv)

∂ K (G\) = ∂ K () for all K , ⊂ G; ∂ K (g) = (∂ K ())g for all K , ⊂ G and g ∈ G; ∂ K (1 ∪ 2 ) ⊂ ∂ K (1 ) ∪ ∂ K (2 ) for all K , 1 , 2 ⊂ G; ∂ K (\A) ⊂ ∂ K () ∪ ∂ K (A) for all K , A, ⊂ G such that A ⊂ .

Proof Let K , , 1 , 2 , A ⊂ G with A ⊂ and let g ∈ G. Property (i) immediately follows from the observation above that ∂ K () consists of all h ∈ G such that K h meets both and G\ since G\(G\) = . Property (ii) follows from the fact that, given an element h ∈ G, the set K h meets both and G\ if and only if the set K hg meets both g and (G\)g = G\g. Every element in ∂ K (1 ∪ 2 ) is in ∂ K (1 ) or in ∂ K (2 ) since G\(1 ∪ 2 ) = (G\1 ) ∩ (G\2 ). This shows (iii). Property (iv) immediately follows from (i) and (iii) above since \A = G\((G\ ) ∪ A). Observe that Int K () ⊂ Cl K () whenever K = ∅. Note also that one always has Cl K () = K −1 . This shows in particular that if the sets K and are finite then Cl K () and ∂ K () are also finite (of cardinality bounded above by |K |||). Definition 9.4.7 Let and K be finite subsets of a group G with = ∅. The relative amenability constant of with respect to K is the rational number α(, K ) ≥ 0 defined by |∂ K ()| . α(, K ) := || Lemma 9.4.8 Let G be a countable amenable group. Let (Fn )n≥1 be a Følner sequence for G and K ⊂ G a finite subset. Then one has lim α(Fn , K ) = 0.

n→∞

Proof First observe that ∂ K (Fn ) = K −1 Fn \ Int K (Fn ) (K −1 Fn \k −1 Fn ) = k∈K

=

h,k∈K

(h −1 Fn \k −1 Fn ).

176

9 Amenable Groups

This implies −1 −1 |∂ K (Fn )| = (h Fn \k Fn ) h,k∈K ≤ |h −1 Fn \k −1 Fn | h,k∈K

=

|Fn \hk −1 Fn |,

h,k∈K

and hence α(Fn , K ) =

|Fn \hk −1 Fn | |∂ K (Fn )| ≤ , |Fn | |Fn | h,k∈K

which shows that α(Fn , K ) tends to 0 as n goes to infinity since |Fn \g Fn | =0 n→∞ |Fn | lim

for all g ∈ G, by definition of a Følner sequence.

Lemma 9.4.9 Let and K be finite subsets of a group G with = ∅. Then one has α(g, K ) = α(, K ) (9.4.3) for all g ∈ G. Proof By using Lemma 9.4.6(ii), we get |∂ K (g)| = |(∂ K ())g| = |∂ K ()|, which yields (9.4.3) after dividing by ||.

Lemma 9.4.10 Let G be a group. Let K be a finite subset of G and 0 < ε < 1. Suppose that (A j ) j∈J is an ε-disjoint finite family of non-empty finite subsets of G. Then one has ⎛ ⎞ 1 max α(A j , K ). α⎝ Aj, K⎠ ≤ 1 − ε j∈J j∈J

Proof Let us set M := max α(A j , K ). j∈J

9.4 The Subadditive Convergence Theorem for Amenable Groups

177

It follows from Lemma 9.4.6(iii) that ⎛ ∂K ⎝

⎞ Aj⎠ ⊂

j∈J

∂ K (A j ).

j∈J

Thus, we have that ⎛ ⎞ ∂ K ⎝ ⎠ ≤ A ∂ (A ) j K j j∈J j∈J ≤ |∂ K (A j )| j∈J

=

α(A j , K )|A j |

j∈J

≤M

|A j |.

j∈J

As the family (A j ) j∈J is ε-disjoint, it then follows from Lemma 9.4.4 that ⎛ α⎝

j∈J

A ∂ K j j∈J M ⎠ ≤ . Aj, K = 1−ε j∈J A j ⎞

Lemma 9.4.11 Let G be a group. Let K , A and be finite subsets of G such that ∅ = A ⊂ . Suppose that ε > 0 is a real number such that |\A| ≥ ε||. Then one has α(, K ) + α(A, K ) . (9.4.4) α(\A, K ) ≤ ε Proof By Lemma 9.4.6(iv), we have the inclusion ∂ K (\A) ⊂ ∂ K () ∪ ∂ K (A).

(9.4.5)

It follows that |∂ K (\A)| |\A| |∂ K (\A)| ≤ ε|| |∂ K () ∪ ∂ K (A)| ≤ ε||

α(\A, K ) =

(since |\A| ≥ ε|| by hypothesis) (by (9.4.5))

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9 Amenable Groups

|∂ K ()| + |∂ K (A)| ε|| ||α(, K ) + |A|α(A, K ) = ε|| α(, K ) + α(A, K ) ≤ ε ≤

(since |A| ≤ ||).

This shows (9.4.4).

Definition 9.4.12 Let G be a group and ε > 0. Let K and be non-empty finite subsets of G. A finite subset P ⊂ G is called an (ε, K )-filling pattern for if the following conditions are satisfied: (FP1) K P ⊂ ; (FP2) the family (K g)g∈P is ε-disjoint. Lemma 9.4.13 Let G be a group and 0 < ε ≤ 1. Let and K be non-empty finite subsets of G. Then there exists an (ε, K )-filling pattern P for such that |K P| ≥ ε(1 − α(, K ))||.

(9.4.6)

Proof Let P denote the set consisting of all (ε, K )-filling patterns for . Observe that P is not empty since ∅ ∈ P. Note also that every element P ∈ P has cardinality bounded above by |||K | since P ⊂ K −1 by (FP1). Choose a pattern P ∈ P with maximal cardinality. Let us show that (9.4.6) is satisfied. By applying Lemma 9.4.2 with A := K and B := K P, we get

|K g ∩ K P| = |K ||K P|.

(9.4.7)

g∈G

We claim that ε|K | ≤ |K g ∩ K P| for all g ∈ Int K ().

(9.4.8)

Indeed, suppose first that g ∈ P. Then we have K g ∩ K P = K g and hence (9.4.8) is satisfied since |K g| = |K | and ε ≤ 1. Suppose now that g ∈ Int K ()\P. If (9.4.8) were not satisfied, then we would have |K g ∩ K P| < ε|K | = ε|K g| and hence P ∪ {g} would be an (ε, K )-filling pattern for , contradicting the maximality of the cardinality of P. This completes the proof of (9.4.8). Finally, we obtain ε|K || Int K ()| =

ε|K |

g∈Int K ()

≤

|K g ∩ K P|

g∈Int K ()

≤

g∈G

|K g ∩ K P|

(by (9.4.8))

9.4 The Subadditive Convergence Theorem for Amenable Groups

= |K ||K P|

179

(by (9.4.7)).

After dividing by |K |, we get ε| Int K ()| ≤ |K P|. This gives us (9.4.6) since (1−α(, K ))|| = ||−|∂ K ()| ≤ |K −1 |−|∂ K ()| = | Cl K ()|−|∂ K ()| = | Int K ()|.

1 Lemma 9.4.14 (Filling lemma) Let G be a group and 0 < ε ≤ . Then there exists 2 an integer s0 = s0 (ε) ≥ 1 such that for each integer s ≥ s0 the following holds. If K 1 , K 2 , . . . , K s are non-empty finite subsets of G such that α(K k , K j ) ≤ ε2s for all 1 ≤ j < k ≤ s,

(9.4.9)

and D is a non-empty finite subset of G such that α(D, K j ) ≤ ε2s for all 1 ≤ j ≤ s,

(9.4.10)

then there exists a sequence P1 , P2 , . . . , Ps of finite subsets of G satisfying the following conditions: (T1) for every 1 ≤ j ≤ s, the set P j is an (ε, K j )-filling pattern for D; (T2) the subsets K j P j ⊂ D, 1 ≤ j ≤ s, are pairwise disjoint; (T3) the subset D ⊂ D defined by D := D\

K j Pj

1≤ j≤s

has cardinality |D | ≤ ε|D|. Proof Fix an integer s ≥ 1. Let K j , 1 ≤ j ≤ s, and D be non-empty finite subsets of G satisfying conditions (9.4.9) and (9.4.10). Let us first describe, by decreasing induction on j, a finite process with at most s steps for constructing finite subsets P j ⊂ G for 1 ≤ j ≤ s. We will see that these subsets have the required properties when s is large enough, namely for s ≥ s0 with s0 = s0 (ε) that will be made precise at the end of the proof. Step 1. We put D0 := D. By (9.4.10), we have α(D0 , K j ) ≤ ε2s for all 1 ≤ j ≤ s. By applying Lemma 9.4.13 with := D0 = D and K = K s , we can find a finite subset Ps ⊂ G such that Ps is an (ε, K s )-filling pattern for D0 and |K s Ps | ≥ ε(1 − α(D, K s ))|D|.

(9.4.11)

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9 Amenable Groups

We deduce from (9.4.11) and (9.4.10) that |K s Ps | ≥ ε(1 − ε2s )|D|.

(9.4.12)

Setting D1 := D0 \K s Ps , we deduce from (9.4.12) that |D1 | ≤ 1 − ε(1 − ε2s ) |D|. Step k. We continue this process by induction in the following way. Suppose that the process has been applied k times, with 1 ≤ k ≤ s − 1. It is assumed that the induction hypotheses at step k are the following: (H(k;a)) Dk−1 is a subset of D satisfying α(Dk−1 , K j ) ≤ (2k − 1)ε2s−k+1 for all 1 ≤ j ≤ s − k + 1; (H(k;b)) Ps−k+1 ⊂ G is an (ε, K s−k+1 )-filling pattern for Dk−1 ; (H(k;c)) setting Dk := Dk−1 \K s−k+1 Ps−k+1 , we have that |Dk | ≤

1 − ε 1 − (2i + 1)ε2s−i |D|.

0≤i≤k−1

Note that these induction hypotheses are satisfied for k = 1 by Step 1. Let us pass from Step k to Step k + 1. Step k + 1. We distinguish two cases. Case 1. Suppose that |Dk | ≤ ε|Dk−1 | and hence |Dk | ≤ ε|D|. Then we take P j = ∅ for all 1 ≤ j ≤ s − k and stop the process. Case 2. Suppose on the contrary that |Dk | > ε|Dk−1 |. Let us first estimate from above, for all 1 ≤ j ≤ s − k, the relative amenability constants α(Dk , K j ). Let 1 ≤ j ≤ s − k. If Ps−k+1 = ∅, then Dk = Dk−1 and therefore α(Dk , K j ) = α(Dk−1 , K j ) ≤ (2k − 1)ε2s−k+1

(by our induction hypothesis (H(k;a)))

≤ (2k + 1)ε2s−k

(since 0 < ε < 1).

Suppose now that Ps−k+1 = ∅. Then we can apply Lemma 9.4.11 with := Dk−1 and A := K s−k+1 Ps−k+1 . This gives us α(Dk , K j ) = α(Dk−1 \K s−k+1 Ps−k+1 , K j ) ≤

α(Dk−1 , K j ) + α(K s−k+1 Ps−k+1 , K j ) . ε

(9.4.13)

9.4 The Subadditive Convergence Theorem for Amenable Groups

181

On the other hand, Lemma 9.4.9 and condition (9.4.9) imply that, for all g ∈ G, α(K s−k+1 g, K j ) = α(K s−k+1 , K j ) ≤ ε2s . As the family (K s−k+1 g)g∈Ps−k+1 is ε-disjoint, this last inequality together with Lemma 9.4.10 give us ⎛ α(K s−k+1 Ps−k+1 , K j ) = α ⎝

⎞ K s−k+1 g, K j ⎠ ≤

g∈Ps−k+1

ε2s . 1−ε

From (9.4.13) and the induction hypothesis (H(k;a)), we deduce that α(Dk , K j ) ≤

(2k − 1)ε2s−k+1 ε2s + ≤ (2k + 1)ε2s−k ε (1 − ε) ε

(for the second inequality, observe that 1/(1 − ε) ≤ 2 since 0 < ε ≤ 1/2 ). This shows (H(k+1;a)). Using Lemma 9.4.13 with := Dk and K := K s−k , we can find a finite subset Ps−k ⊂ G such that Ps−k is an (ε, K s−k )-filling pattern for Dk , thus yielding (H(k+1;b)), and satisfying |K s−k Ps−k | ≥ ε 1 − α(Dk , K s−k ) |Dk | ≥ ε 1 − (2k + 1)ε2s−k |Dk |. (9.4.14) Setting Dk+1 := Dk \K s−k Ps−k , we deduce from (9.4.14) that |Dk+1 | ≤ |Dk | 1 − ε 1 − (2k + 1)ε2s−k . Together with the inequality of the induction hypothesis (H(k;c)), this yields |Dk+1 | ≤ |D|

1 − ε 1 − (2i + 1)ε2s−i . 0≤i≤k

Thus condition (H(k+1;c)) is also satisfied. This finishes the construction of Step k + 1 and proves the induction step. Now, suppose that this process continues until Step s. Using (H(k;c)) for k = s, we obtain 1 − ε 1 − (2i + 1)ε2s−i . |Ds | ≤ |D| (9.4.15) 0≤i≤s−1

To conclude, let us show that for s ≥ s0 , with s0 = s0 (ε) depending only on ε, we have |Ds | ≤ ε|D|.

182

9 Amenable Groups

As (2i + 1)ε2s−i ≤ (2s + 1)εs+1 for all 0 ≤ i ≤ s − 1, we deduce from (9.4.15) that s (9.4.16) |Ds | ≤ |D| 1 − ε(1 − (2s + 1)εs+1 ) . Since limr →+∞ (2r + 1)εr +1 = 0 and limr →+∞ (1 − 2ε )r = 0, both monotonically for large r , we can find an integer s0 = s0 (ε) ≥ 1 such that for all r ≥ s0 , we have both (2r + 1)εr +1 ≤ 21 and (1 − 2ε )r ≤ ε. Now, if s ≥ s0 , using inequality (9.4.16) we deduce that ε s ≤ ε|D|. |Ds | ≤ |D| 1 − 2

This completes the proof of the lemma.

Proof of Theorem 9.4.1 Let A ∈ P f in (G). By taking A = B in (H1), we get h(A) ≤ 2h(A) and hence 0 ≤ h(A). (9.4.17) On the other hand, we have that ⎛ ⎞ h(A) = h ⎝ {g}⎠ ≤

g∈A

h({g})

(by (H1))

g∈A

= h({1G })|A| so that, setting M := h({1G }),

(by (H2)),

h(A) ≤ M|A|.

(9.4.18)

Let (Fn )n≥1 be a Følner sequence for G. By Lemma 9.4.8, we know that lim α(Fn , K ) = 0 for every finite subset K ⊂ G.

n→∞

(9.4.19)

Consider the infimum limit λ := lim inf n→∞

h(Fn ) . |Fn |

(9.4.20)

Note that 0 ≤ λ ≤ M by (9.4.17) and (9.4.18). 1 Let 0 < ε ≤ and choose an integer s ≥ s0 , where s0 = s0 (ε) is as in 2 Lemma 9.4.14. Recall that one says that a finite sequence (K j )1≤ j≤s is extracted from the sequence (Fn )n≥1 if there are positive integers n 1 < n 2 < · · · < n s such that K j = Fn j for all 1 ≤ j ≤ s.

9.4 The Subadditive Convergence Theorem for Amenable Groups

183

It clearly follows from (9.4.19) and (9.4.20) that we can find, by induction on j, a finite sequence (K j )1≤ j≤s extracted from the sequence (Fn )n≥1 such that α(K k , K j ) ≤ ε2s for all 1 ≤ j < k ≤ s and

h(K j ) ≤ λ + ε for all 1 ≤ j ≤ s. |K j |

(9.4.21)

Let D ⊂ G be a non-empty finite subset satisfying α(D, K j ) ≤ ε2s for all 1 ≤ j ≤ s. By Lemma 9.4.14, we can find a sequence (P j )1≤ j≤s of finite subsets of G satisfying the following conditions: (T1) the set P j is an (ε, K j )-filling pattern for D for every 1 ≤ j ≤ s; (T2) the subsets K j P j ⊂ D, 1 ≤ j ≤ s, are pairwise disjoint; (T3) the subset D ⊂ D defined by

D := D\

K j Pj

1≤ j≤s

is such that |D | ≤ ε|D|. ⎛

As

D = D ∪ ⎝

⎞ K j Pj ⎠ ,

1≤ j≤s

it follows from the subadditivity property (H1) that h(D) ≤ h(D ) +

h(K j P j ).

(9.4.22)

1≤ j≤s

Now, since |D | ≤ ε|D| by (T3), we deduce from (9.4.18) that h(D ) ≤ Mε|D|.

(9.4.23)

On the other hand, for all 1 ≤ j ≤ s, we have that ⎛

h(K j P j ) = h ⎝

⎞ K j g⎠

g∈P j

≤

g∈P j

h(K j g)

(by the subadditivity property (H1))

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9 Amenable Groups

=

h(K j )

(by the right-invariance property (H2))

g∈P j

h(K j ) |K j g| |K j | g∈P j |K j g| ≤ (λ + ε) =

(since |K j | = |K j g|) (by (9.4.21)).

g∈P j

As the family (K j g)g∈P j is ε-disjoint by (T1), we then deduce from Lemma 9.4.4 that λ+ε λ+ε |K j P j |. h(K j P j ) ≤ K j g = 1−ε 1−ε g∈P j This gives us

h(K j P j ) ≤

1≤ j≤s

and hence

λ+ε |K j P j | 1−ε 1≤ j≤s

h(K j P j ) ≤

1≤ j≤s

λ+ε |D|, 1−ε

(9.4.24)

since the sets K j P j , 1 ≤ j ≤ s, are pairwise disjoint subsets of D by (T2). Combining (9.4.22)–(9.4.24), we get λ+ε h(D) ≤ Mε + . |D| 1−ε By (9.4.19), we can find an integer n 0 ≥ 1 such that, for all n ≥ n 0 , α(Fn , K j ) ≤ ε2s for all 1 ≤ j ≤ s. By replacing D by Fn in (9.4.25), we obtain λ+ε h(Fn ) ≤ Mε + |Fn | 1−ε for all n ≥ n 0 . By letting n tend to infinity, this gives us lim sup n→∞

λ+ε h(Fn ) ≤ + Mε. |Fn | 1−ε

(9.4.25)

9.4 The Subadditive Convergence Theorem for Amenable Groups

185

Letting now ε tend to 0, we get lim sup n→∞

h(Fn ) ≤ λ. |Fn |

Consequently, we have that λ = lim inf n→∞

This proves that the sequence

h(Fn ) h(Fn ) = lim sup . |Fn | n→∞ |Fn |

h(Fn ) |Fn |

converges to λ. n≥1

h(Fn ) does not depend on the choice |Fn | of the Følner sequence (Fn )n≥1 . To see this, suppose that (Fn )n≥1 is another Følner h(Fn ) . sequence for G and let λ := limn→∞ |Fn | Consider the sequence (E n )n≥1 defined by It only remains to show that λ = limn→∞

E n :=

Fn if n is odd Fn if n is even.

Clearly (E n )n≥1 is also a Følner sequence. Therefore, the sequence has a limit μ. We get μ = λ = λ by uniqueness of the limit.

h(E n ) |E n |

n≥1

Notes A detailed exposition of the theory of amenable groups may be found for example in [41, 85, 88], and [22, Chap. 4]. The notes by Tao [103] provide an especially nice introduction to amenability via Følner sequences. Amenability theory has its roots in the difficulties raised at the beginning of the 20th century by both the definition of the Lebesgue integral and the Banach-Tarski paradox (see [27] for a historical survey). Amenable groups were introduced by von Neumann [113] in 1929. The original definition of von Neumann requires that the group admits an invariant finitely-additive probability measure defined on the set of all of its subsets. A fundamental observation of M. Day is that von Neumann’s definition is equivalent to the existence of an invariant mean on the space of bounded functions on the group. The introduction of means has the advantage of allowing the use of the powerful tools of functional analysis. Let G be a group. For g ∈ G, we denote by L g and Rg the left and right multiplication by g, that is, the maps L g : G → G and Rg : G → G defined by L g (h) = gh and Rg (h) = hg for all h ∈ G. Consider

186

9 Amenable Groups

the vector space ∞ (G) consisting of all bounded real-valued maps f : G → R. A mean on G is a linear map m : ∞ (G) → R such that inf g∈G f (g) ≤ m( f ) ≤ supg∈G f (g) for all f ∈ ∞ (G). One says that a mean m on G is left-invariant (resp. right-invariant) if it satisfies m( f ◦ L g ) = m( f ) (resp. m( f ◦ Rg ) = m( f )) for all f ∈ ∞ (G) and g ∈ G. A mean m on G is said to be bi-invariant if it is both left and right-invariant. The following conditions are all equivalent: (1) G is amenable; (2) G admits a left-invariant mean; (3) G admits a right-invariant mean; (4) G admits a bi-invariant mean. There is a natural one-to-one correspondence between means on a group G and finitely-additive probability measures on subsets of G which is given by m → μm , where μm (A) is the value taken by m at the characteristic map χ A of A ⊂ G. This correspondence between means and finitely-additive probability measures preserves left and right-invariance and explains the equivalence between von Neumann’s and Day’s definitions. In fact, many other equivalent definitions of amenability for groups may be found in the literature and a complete list, if it exists, would be certainly far too long to be given here. The interest of choosing one of these definitions rather than another depends on the context. It seems that the term “amenable” was used for the first time by M. Day in 1949. The German word originally used by von Neumann in 1929 was “messbar”. Note that “amenable” is an anagram of “meanable” and that the French word that is currently used for “amenable” is “moyennable”. The Tits alternative [106] asserts that every finitely generated linear group either contains a non-abelian free subgroup or is virtually solvable (a group G is called linear if there exist a field K and an integer n ≥ 1 such that G is isomorphic to a subgroup of GLn (K )). One deduces from the Tits alternative that if G is a linear group then the following conditions are equivalent: (1) G is amenable; (2) G is locally virtually solvable; (3) G contains no non-abelian free subgroups. The group GL2 (K ), where K is the algebraic closure of a finite field, provides an example of a linear group that is locally virtually solvable (it is even locally finite, see Exercise 9.10) but not virtually solvable. However, in characteristic 0, every linear group that is locally virtually solvable is virtually solvable. There exist finitely generated amenable groups that are not virtually solvable. The groups of intermediate growth, e.g., Grigorchuk groups [42], are examples of such groups. The infinite finitely generated amenable simple groups exhibited by Juschenko and Monod in [52] provide further examples of finitely generated amenable groups that are not virtually solvable, since it is clear that an infinite simple group cannot be virtually solvable. On the other hand, there exist non-amenable groups that contain no non-abelian free subgroups among Tarski monsters [82] and free Burnside groups [2]. These examples illustrate the fact that both the class of amenable groups and the class of non-amenable groups are very hard to apprehend from an algebraic viewpoint. A net in a set X is a family (xi )i∈I of elements of X indexed by a directed set I (recall that a directed set is a partially ordered set (I, ≤) satisfying the following condition: for all i 1 , i 2 ∈ I , there exists j ∈ I such that i 1 ≤ j and i 2 ≤ j). In a topological space, the notion of a limit can be extended to nets of points. For example, one says that a net (xi )i∈I of real numbers converges to 0, and one writes limi xi = 0, if, for every ε > 0, there exists i 0 ∈ I such that |xi | ≤ ε for all i 0 ≤ i. If G is a

Notes

187

group, a Følner net for G is a net (Fi )i∈I of non-empty finite subsets of G such that limi |Fi \g Fi | = 0 for all g ∈ G. By adapting the proof of Lemma 9.2.1, one easily shows that a (possibly uncountable) group is amenable if and only if it admits a Følner net. On the other hand, given a Følner net (Fi )i∈I , there is an associated net of means (m i )i∈I , where m i ( f ) is the average of f ∈ ∞ (G) on Fi . Then, by the compactness of the closed unit ball of the dual space of ∞ (G) for the weak- topology (which follows from the Banach-Alaoglu theorem), there exists a subnet of (m i )i∈I that converges to a mean m. From the fact that (Fi )i∈I is a Følner net, one easily deduces that the limit mean m is left-invariant. The converse implication, namely that the existence of a left-invariant mean implies the existence of a Følner net, is more delicate. The proof given by Følner [37] for this converse implication was subsequently simplified by Namioka [81]. The definition of amenability via the existence of invariant means may be extended to semigroups, i.e., sets equipped with a binary operation that is only assumed to be associative, but theoretical complications appear in this more general setting (see [25, 26, 81, 85]). For instance, when considering semigroups, one must distinguish between left-amenability and right-amenability. Also, no equivalent definition of amenability based on Følner-type conditions is available in this setting. However, for semigroups, there is a Følner-type condition that is implied by amenability and a stronger Følner-type condition that implies amenability. The notion of amenability has been generalized in several other directions (group actions, groupoids, associative algebras, orbit equivalences, etc.) and plays now an important role in many branches of mathematics such as combinatorial and geometric group theory, ergodic theory, dynamical systems, geometry of manifolds, and operator algebras. This is due to the fact that amenable objects are easier to manipulate because they are close to finite and commutative ones. The Baumslag-Solitar groups B S(m, n) were introduced in [12]. Given non-zero integers m, n, the Baumslag-Solitar group B S(m, n) is the group given by the presentation a, b : ba m b−1 = a n . This means that B S(m, n) is the quotient of the free group F on two generators a and b by the smallest normal subgroup N of F such that ba m b−1 a −n ∈ N . Baumslag-Solitar groups are often used as counterexamples in combinatorial and geometric group theory. For instance, the Baumslag-Solitar group B S(2, 3) is non-Hopfian (a group G is called Hopfian if every surjective endomorphism of G is injective). In contrast, it follows from results due to Malcev that every finitely generated linear group is residually finite and that every finitely generated residually finite group is Hopfian (cf. Exercise 10.8 for the definition of residual finiteness). For a nice survey on Baumslag-Solitar groups, see [1]. A proof of Theorem 9.4.1, under the additional hypothesis that h is nondecreasing, was given by Lindenstrauss and Weiss in [74, Theorem 6.1]. Their proof is based on the theory of quasi-tiles in amenable groups that was developed by Ornstein and Weiss in [83]. An alternative proof of Theorem 9.4.1 was sketched by Gromov [44] (see [61] for a detailed exposition of Gromov’s argument). A version of Theorem 9.4.1 for cancellative one-sided amenable semigroups was given in [23]. The proof of Theorem 9.4.1 presented in the present chapter relies on Gromov’s

188

9 Amenable Groups

ideas and closely follows the exposition that may be found in [23]. The extension to uncountable amenable groups, for which Følner sequences are replaced by Følner nets, is straightforward (cf. [23]).

Exercises 9.1 Show that the additive group Q of rational numbers is countable but not finitely generated. 9.2 Show that the sequence (Fn )n≥1 , where Fn :=

k | k ∈ N and k ≤ (n + 1)! n!

for all n ≥ 1, is a Følner sequence for Q. 9.3 The symmetric difference of two sets A and B is the set A B consisting of all elements that are either in A or in B but not in both. Thus one has A B = A ∪ B\A ∩ B. (a) Show that one has A B = (A\B) ∪ (B\A). (b) Show that one has A B = ∅ if and only if A = B. (c) Show that if A and B are finite sets with the same cardinality then one has |A B| = 2|A\B| = 2|B\A|. (d) Let G be a group and let (Fn )n≥1 be a sequence of non-empty finite subsets of G. Show that (Fn )n≥1 is a Følner sequence for G if and only if one has |Fn g Fn | limn→∞ = 0 for all g ∈ G. |Fn | 9.4 Let G be a countable amenable group. Let (Fn )n≥1 be a Følner sequence for G and let (gn )n≥1 be a sequence of elements of G. Show that (Fn gn )n≥1 is a Følner sequence for G. 9.5 Let G be a group and let A be a finite subset of G. Show that if (Fn )n≥1 is a Følner sequence for G then (Fn ∪ A)n≥1 is also a Følner sequence for G. 9.6 One says that a Følner sequence (Fn )n≥1 for a group G is a Følner exhaustion if it satisfies Fn ⊂ Fn+1 for all n ≥ 1 and G = n≥1 Fn . Show that every countable amenable group G admits a Følner exhaustion. 9.7 Show that any group G satisfies the following condition: for every s ∈ G and every ε > 0, there exists a non-empty finite subset F ⊂ G such that |F\s F| ≤ ε|F|. 9.8 Deduce from Lemma 9.2.12 that a group G is amenable if and only if it satisfies the following condition: for every finite subset S of G and every ε > 0, there exists a non-empty finite subset F ⊂ G such that |S F| ≤ (1 + ε)|F|. Hint: observe that S F ⊂ F ∪ S F and S F\F = (S ∪ {1G })F\F for all F, S ⊂ G. 9.9 Let G be a group.

Exercises

189

(a) Show that the following conditions are equivalent: (1) G is countable and locally finite; (2) there exists a non-decreasing sequence (Hn )n≥1 of finite subgroups of G such that G = n≥1 Hn . (b) Suppose that G is countable and locally finite. Let (Hn )n≥1 be a nondecreasing sequence of finite subgroups of G such that G = n≥1 Hn . Show that the sequence (Hn )n≥1 is a Følner exhaustion of G (cf. Exercise 9.6). 9.10 Let K be the algebraic closure of a finite field and let n be a positive integer. Show that the group GLn (K ) is locally finite. Hint: Observe that K is the union of an increasing sequence of finite subfields. 9.11 Let G be a group. Suppose that G contains a normal subgroup H such that H is solvable of solvability degree m and G/H is solvable of solvability degree n. Show that G is solvable of solvability degree at most m + n. 9.12 Let G be a group and let (D n (G))n≥0 denote its derived series. Show that D n (G) is normal in G for every n ≥ 0. 9.13 Show that every virtually solvable group that is a torsion group is locally finite. Hint: reduce to the case of a solvable group and then use induction on the solvability degree. 9.14 Let G be a finitely generated group. Suppose that A ⊂ G is a finite generating subset of G. Show that the following conditions are all equivalent: (1) G is amenable; (2) for every ε > 0, there exists a non-empty finite subset F ⊂ G such that |F\a F| ≤ ε|F| for all a ∈ A; (3) for every ε > 0, there exists a non-empty finite subset F ⊂ G such that |AF\F| ≤ ε|F|; (4) for every ε > 0, there exists a non-empty finite subset F ⊂ G such that |AF| ≤ (1 + ε)|F|. 9.15 Show that a group G is amenable if and only if it satisfies the following condition: for every finite subset K ⊂ G and every real number ε > 0, there exists a non-empty finite subset F ⊂ G such that α(F, K ) ≤ ε. 9.16 (Groups with subexponential growth). Let G be a finitely generated group. Let A ⊂ G be a finite subset which generates G as a semigroup (i.e., every element of G can be written as a finite product of elements of A). For n ≥ 1, let Bn = Bn (G, A) denote the set consisting of all g ∈ G that can be written in the form g = a1 a2 . . . ak with 0 ≤ k ≤ n and ai ∈ A for all 1 ≤ i ≤ k. (a) Show that the limit λ = λ(G, A) := lim

n→∞

log |Bn | n

exists and that 0 ≤ λ < ∞. Hint: see Exercise 6.4. (b) One says that G has subexponential growth if λ = 0 and that G has exponential growth otherwise. Show that this definition does not depend on the choice of the finite subset A ⊂ G which generates G as a semigroup. Hint: Observe that if A is another finite subset of G which generates G as a semigroup and Bn := Bn (G, A ), then there exists a positive integer

for all n ≥ 1. C = C(G, A, A ) such that Bn ⊂ BCn

190

9 Amenable Groups

(c) Show that if G has subexponential growth then lim inf n→∞

|Bn+1 | = 1. |Bn |

(d) Show that if G has subexponential growth then G is amenable. Hint: Consider a finite subset S ⊂ G and ε > 0. Choose a finite subset A ⊂ G that generates G as a semigroup with S ⊂ A. Observe that the subsets Bn := Bn (G, A) satisfy S Bn ⊂ Bn+1 for all n ≥ 1 and deduce from the result of the previous question that if G has subexponential growth then there exists n ≥ 1 such that |Bn+1 \Bn | ≤ ε|Bn |. Conclude by using the characterization of amenability in Lemma 9.2.12. 9.17 Let G be a finitely generated group and H a subgroup of G. (a) Show that if H is finitely generated and G has subexponential growth, then H has subexponential growth. (b) Suppose that H is of finite index in G. Show that H is finitely generated and that G has subexponential growth if and only if H has subexponential growth. 9.18 Show that every finitely generated abelian group has subexponential growth. 9.19 Show that the integral Heisenberg group HZ described in Example 9.2.19 is finitely generated and has subexponential growth. Hint: Observe that the matrices X := M(1, 0, 0), Y := M(0, 1, 0), and Z := M(0, 0, 1) generate the group HZ and that they satisfy [X, Y ] = Z and [X, Z ] = [Y, Z ] = 1. 9.20 (An example of a finitely generated amenable group with exponential growth). Consider the Baumslag-Solitar group G := B S(1, 2), i.e., the group of affine transformations of the real line generated by the translation t : x → x + 1 and the homothety h : x → 2x (cf. Example 9.2.18). We have seen in Example 9.2.18 that G is metabelian. Therefore G is amenable by Corollary 9.2.16. The goal of this exercise is to show that G has exponential growth. Let a and b be the elements of G respectively defined by a := h −1 and b := at. (a) Show that if u 1 , . . . , u n and v1 , . . . , vm are two finite sequences of elements of {a, b} such that u 1 . . . u n = v1 . . . vm then n = m and u i = vi for all 1 ≤ i ≤ n. Hint: use a ping-pong-type argument after observing that a sends the open interval (0, 1) in (0, 1/2) and that b sends (0, 1) in (1/2, 1). (b) Deduce from the result of the previous question that G has exponential growth. (c) Find a similar argument to prove that B S(1, n) has exponential growth for every n ≥ 2.

Chapter 10

Mean Topological Dimension for Actions of Amenable Groups

In this chapter, by a “dynamical system”, we mean a triple (X, G, T ), where X is a topological space, G a group, and T : G × X → X a continuous action of G on X (see Sect. 10.1). The mean topological dimension of a dynamical system (X, G, T ), where X is a normal space and G is a countable amenable group, is defined in Sect. 10.2. In the case G = Z, it coincides with the mean topological dimension of the time 1 homeomorphism associated with the action. We extend most of the results of Chaps. 6 and 7. We prove in particular that if G is a countably-infinite amenable group and P is a polyhedron, then the mean topological dimension of the G-shift on P G is equal to dim(P) (Corollary 10.6.3). We also show that if G is a countable amenable group admitting subgroups of arbitrarily large finite index and P is a polyhedron, then the mean topological dimension of closed invariant subspaces of K G take all real values in the interval [0, dim(P)] (Theorem 10.8.1).

10.1 Continuous Actions Let G be a group. For our exposition, we shall use a multiplicative notation for the binary operation on G. An action of the group G on a set X is a map T : G × X → X satisfying T (g, T (h, x)) = T (gh, x) and T (1G , x) = x for all g, h ∈ G and x ∈ X (here 1G is the identity element of G). Denoting by Tg : X → X the map defined by Tg (x) = T (g, x) for all x ∈ X , this amounts to saying that one has (Act-1) Tg ◦ Th = Tgh for all g, h ∈ G, and (Act-2) T1G = Id X , where Id X is the identity map on X . It follows from (Act-1) and (Act-2) that Tg is bijective for all g ∈ G with inverse map Tg−1 := (Tg )−1 = Tg−1 . Let X be a set equipped with an action T : G × X → X of a group G. One says that a subset Y ⊂ X is T -invariant if Y is Tg -invariant, i.e., Tg (Y ) ⊂ Y , for all g ∈ G. Note that if Y ⊂ X is T -invariant then one actually has Tg (Y ) = Y for all g ∈ G. Indeed, we then have on one hand Tg (Y ) ⊂ Y (since Y is Tg -invariant) and on the other one Y = Tg (Tg−1 (Y )) ⊂ Tg (Y ) (since Y is Tg−1 -invariant). © Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4_10

191

192

10 Mean Topological Dimension for Actions of Amenable Groups

If Y ⊂ X is T -invariant, then T induces by restriction an action T |Y : G ×Y → Y given by T |Y (g, y) = T (g, y) for all g ∈ G and y ∈ Y . It is a common abuse to write T instead of T |Y if there is no risk of confusion. When X is a topological space, one says that an action T of G on X is continuous if the maps Tg are continuous on X for all g ∈ G. This implies that, for every g ∈ G, the map Tg is a homeomorphism of X with inverse homeomorphism Tg−1 = Tg−1 . In this chapter, by a dynamical system, we shall mean a triple (X, G, T ), where X is a topological space, which is called the phase space of the dynamical system, G is a group, and T : G × X → X is a continuous action of G on X . All dynamical systems will be implicitly assumed to have non-empty phase spaces. Example 10.1.1 Suppose that f : X → X is a homeomorphism of a topological space X . Then the map T : Z × X → X , defined by T (n, x) = f n (x) for all n ∈ Z and x ∈ X , is a continuous action of the additive group Z of integers on X . One says that (X, Z, T ) is the dynamical system generated by f . Conversely, if T : Z × X → X is a continuous action of Z on the topological space X , then (X, Z, T ) is generated by the homeomorphism T1 : X → X . Thus, there is a canonical bijection between the set of homeomorphisms of X and the set of continuous actions of Z on X . If T and S are continuous actions of the same group G on two topological spaces X and Y , one says that the dynamical systems (X, G, T ) and (Y, G, S) are topologically conjugate if there exists a homeomorphism ϕ : X → Y such that ϕ ◦ Tg = Sg ◦ ϕ for all g ∈ G. One then says that the homeomorphism ϕ conjugates the dynamical systems (X, G, T ) and (Y, G, S). One says that the dynamical system (X, G, T ) embeds in the dynamical system (Y, G, S) if there exists a topological embedding f : X → Y such that f ◦ Tg = Sg ◦ f for all g ∈ G. One then says that f is an embedding of the dynamical system (X, G, T ) in the dynamical system (Y, G, S). Note that the system (X, G, T ) embeds in (Y, G, S) if and only if there exists an Sinvariant subset Z ⊂ Y such that (X, G, T ) is topologically conjugate to (Z , G, S). Example 10.1.2 Let X and Y be topological spaces. Suppose that f is a homeomorphism of X and g is a homeomorphism of Y . Let (X, Z, T ) and (Y, Z, S) denote the dynamical systems generated by f and g respectively. Then (X, Z, T ) and (Y, Z, S) are topologically conjugate if and only if the homeomorphisms f and g are topologically conjugate. Moreover, a homeomorphism ϕ : X → Y conjugates (X, Z, T ) and (Y, Z, S) if and only if it conjugates f and g. Similarly, (X, Z, T ) embeds in (Y, Z, S) if and only if (X, f ) embeds in (Y, g).

10.2 Definition of Mean Topological Dimension Let (X, G, T ) be a dynamical system, i.e., a topological space X equipped with a continuous action T : G × X → X of a group G. Let P f in (G) denote the set consisting of all finite subsets of G.

10.2 Definition of Mean Topological Dimension

193

Given a finite open cover α = (Ui )i∈I of X and an element A ∈ P f in (G), we define the finite open cover α A = α A (X, G, T ) by α A :=

Tg−1 (α).

(10.2.1)

g∈A

Formally, α A is the family indexed by I A (the set consisting of all maps from A to I ) formed by all the open subsets

Tg−1 (Uι(g) ) ⊂ X,

g∈A

where ι : A → I runs over I A . Proposition 10.2.1 Let X be a topological space equipped with a continuous action T : G × X → X of a group G. Let α be a finite open cover of X . Then the following hold: (i) D(α Ag ) = D(α A ) for all g ∈ G and A ∈ P f in (G); (ii) D(α A ) ≤ D(α B ) for all A, B ∈ P f in (G) such that A ⊂ B; (iii) if X is normal then one has D(α A∪B ) ≤ D(α A ) + D(α B ) for all A, B ∈ P f in (G). Proof If A ∈ P f in (G) and g ∈ G, then α Ag =

−1 Thg (α)

h∈A

=

(Th ◦ Tg )−1 (α)

h∈A

=

Tg−1 ◦ Th−1 (α)

h∈A

=

Tg−1 (Th−1 (α))

h∈A

=

Tg−1

Th−1 (α)

(by Proposition 6.1.1)

h∈A

= Tg−1 (α A ). This shows that the homeomorphism Tg−1 sends α Ag to α A and hence that D(α Ag ) = D(α A ).

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10 Mean Topological Dimension for Actions of Amenable Groups

If A, B ∈ P f in (G) and A ⊂ B, then α B is finer than α A . It follows that D(α A ) ≤ D(α B ) by Proposition 1.1.4. It remains only to establish (iii). So let us assume that X is normal and let A, B ∈ P f in (G). In the particular case when A and B are disjoint, we have that α A∪B = α A ∨ α B and hence D(α A∪B ) = D(α A ∨ α B ) ≤ D(α A ) + D(α B ) by Proposition 6.1.5. For the general case, it suffices to observe that D(α A∪B ) = D(α(A\B)∪B ) ≤ D(α A\B ) + D(α B ) (since A\B and B are disjoint) ≤ D(α A ) + D(α B ), where the last inequality follows from the fact that A\B ⊂ A so that D(α A\B ) ≤ D(α A ) by (ii). Suppose that X is a normal space equipped with a continuous action T : G × X → X of a countable amenable group G. Let α be a finite open cover of X . It follows from assertions (i) and (iii) in Proposition 10.2.1 that the map h : P f in (G) → N defined by h(A) = D(α A ) is right-invariant (i.e., h(Ag) = h(A) for all A ∈ P f in (G) and g ∈ G) and subadditive (i.e., h(A ∪ B) ≤ h(A) + h(B) for all A, B ∈ P f in (G)). Thus, we deduce from Theorem 9.4.1 that if (Fn )n≥1 is a Følner sequence for G, then the limit D(α, X, G, T ) := lim

n→∞

D(α Fn ) , |Fn |

(10.2.2)

exists, is finite, and does not depend on the choice of the Følner sequence (Fn ). Definition 10.2.2 Let X be a normal space equipped with a continuous action T : G × X → X of a countable amenable group G. The mean topological dimension of the dynamical system (X, G, T ) is the quantity 0 ≤ mdim(X, G, T ) ≤ ∞ defined by mdim(X, G, T ) := sup D(α, X, G, T ), α

where α runs over all finite open covers of X and D(α, X, G, T ) is the non-negative real number defined by (10.2.2). Example 10.2.3 Let f be a homeomorphism of a normal space X and consider the dynamical system (X, Z, T ) generated by f . Let us choose the Følner sequence

10.2 Definition of Mean Topological Dimension

195

(Fn )n≥1 for Z given by Fn := {0, 1, . . . , n − 1}. Then, for every finite open cover α of X , we have that α Fn =

n−1

Tk−1 (α) =

k=0

n−1

f −k (α)

k=0

and hence, using the notation introduced in Sect. 6.3, α Fn = ω(α, f, n). It follows that D(α, X, Z, T ) = lim

n→∞

D(α Fn ) D(α, f, n) = lim = D(α, f ). n→∞ |Fn | n

Since, by definition, mdim(X, Z, T ) = sup D(α, X, Z, T ) and mdim(X, f ) = sup D(α, f ), α

α

where α runs over all finite open covers of X , we conclude that mdim(X, Z, T ) = mdim(X, f ).

10.3 General Properties of Mean Topological Dimension Mean topological dimension is an invariant of topological conjugacy. More precisely, we have the following statement: Proposition 10.3.1 Let X and Y be normal spaces equipped with continuous actions T : G × X → X and S : G ×Y → Y of a countable amenable group G. Suppose that the dynamical systems (X, G, T ) and (Y, G, S) are topologically conjugate. Then one has mdim(X, G, T ) = mdim(Y, G, S). Proof Let ϕ : X → Y be a homeomorphism that conjugates the systems (X, G, T ) and (Y, G, S). Let β be a finite open cover of Y and α := ϕ −1 (β). As ϕ ◦ Tg = Sg ◦ ϕ for all g ∈ G, the homeomorphism ϕ sends α A (X, G, T ) to β A (Y, G, S) for every A ∈ P f in (G). It follows that D(α A (X, G, T )) = D(β A (Y, G, S)) for every A ∈ P f in (G). Replacing A by Fn , where (Fn )n≥1 is a Følner sequence for G, we deduce from (10.2.2) that D(α, X, G, T ) = D(β, Y, G, S). Since β → α provides a bijective correspondence between the finite open covers of Y and those of X , we deduce that mdim(X, G, T ) = sup D(α, X, G, T ) = sup D(β, Y, G, S) = mdim(Y, G, S). α

β

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When the phase space of a dynamical system is finite-dimensional, its mean topological dimension is zero. More precisely, we have the following result. Proposition 10.3.2 Let X be a normal space equipped with a continuous action T : G × X → X of a countably-infinite amenable group G. Suppose that dim(X ) < ∞. Then one has mdim(X, G, T ) = 0. Let us start by establishing a general property of Følner sequences in infinite groups. Lemma 10.3.3 Let G be a countably-infinite amenable group and let (Fn )n≥1 be a Følner sequence for G. Then one has lim |Fn | = ∞.

n→∞

(10.3.1)

Proof Let M be a positive integer. As G is infinite, we can find a finite subset S ⊂ G with |S| ≥ M 2 . Since (Fn )n≥1 is a Følner sequence, there is an integer n 0 ≥ 1 such that |Fn \g Fn | ≤ |Fn |/2 for all g ∈ S and n ≥ n 0 . As Fn = ∅, this implies that the set Fn meets g Fn and hence that S is contained in the set An := {x y −1 : x, y ∈ Fn }. It follows that |S| ≤ |Fn |2 and hence that |Fn | ≥ M for all n ≥ n 0 . This shows 10.6. Proof of Proposition 10.3.2 Choose a Følner sequence (Fn )n≥1 for G. For each finite open cover α of X and all A ∈ P f in (G), we have that D(α A ) ≤ dim(X ) by definition of dim(X ). As dim(X ) < ∞ and limn→∞ |Fn | = ∞ by Lemma 10.3.3, we deduce that D(α Fn ) = 0. D(α, X, G, T ) = lim n→∞ |Fn | Thus, we have that mdim(X, G, T ) = sup D(α, X, G, T ) = 0. α

Proposition 10.3.4 Let X be a normal space equipped with a continuous action T : G × X → X of a countable amenable group G. Let Y ⊂ X be a closed (and hence normal) T -invariant subset of X . Then one has mdim(Y, G, T |Y ) ≤ mdim(X, G, T ). Proof Let α = (Ui )i∈I be a finite open cover of Y . For each i ∈ I , we can find an open subset Vi of X such that Ui = Vi ∩ Y . Consider the finite open cover β of X defined

10.3 General Properties of Mean Topological Dimension

197

by β := (Vi )i∈I ∪ (X \Y ). Now let A ∈ P f in (G). Let γ = (W j ) j∈J be a finite open cover of X that is finer than β A (X, G, T ). Then γ := (W j ∩ Y ) j∈J is clearly a finite open cover of Y that is finer than α A (Y, G, T |Y ) and we have ord(γ ) ≤ ord(γ ) (cf. the proof of Proposition 1.2.1). It follows that D(α A (Y, G, T |Y )) ≤ D(β A (X, G, T )) for all A ∈ P f in (G). Replacing A by Fn , where (Fn )n≥1 is a Følner sequence for G, and passing to the limit, we deduce that D(α, Y, G, T |Y ) ≤ D(β, X, G, T ) ≤ mdim(X, G, T ). This implies that mdim(Y, G, T |Y ) = supα D(α, Y, G, T |Y ) ≤ mdim(X, G, T ). Corollary 10.3.5 Let X be a compact space and Y a normal Hausdorff space equipped with continuous actions T : G × X → X and S : G × Y → Y of a countable amenable group G. Suppose that the dynamical system (X, G, T ) embeds in the dynamical system (Y, G, S). Then one has mdim(X, G, T ) ≤ mdim(Y, G, S). Proof Let f : X → Y be a topological embedding of (X, G, T ) in (Y, G, S). It follows from our hypotheses that Z := f (X ) is a closed S-invariant subset of Y and that f induces a topological conjugacy between (X, G, T ) and (Z , G, S| Z ). Thus, we get mdim(X, G, T ) = mdim(Z , G, S| Z ) ≤ mdim(Y, G, S) by applying Propositions 10.3.1 and 10.3.4.

Let H be a subgroup of a group G. Recall that a subset C ⊂ G is called a leftcoset of H if C = g H for some g ∈ G. The left-cosets of H form a partition of G. We denote by G/H the set consisting of all left-cosets of H . Left-multiplication by elements of G induces an action of G on G/H . This action is defined by the map from G × G/H to G/H given by (g, C) → gC for g ∈ G and C ∈ G/H . A subset R ⊂ G is called a complete set of representatives of the left-cosets of H if each left-coset of H contains a unique element belonging to R. Thus, if R is a complete set of representatives of the left-cosets of H , then the cardinality of R is equal to that of G/H , i.e., to the index [G : H ] of H in G. Recall from Proposition 9.2.13 and Corollary 9.2.22 that if H is of finite index in G, then the group G is amenable if and only if H is amenable. If T : G × X → X is a continuous action of a group G on a topological space X and H is a subgroup of G, then T induces by restriction a continuous action T (H ) of H on X defined by T (H ) (h, x) := T (h, x) for all h ∈ H and x ∈ X . When a countable amenable group acts on a normal space, the mean topological dimension of the restriction of the action to a subgroup of finite index is proportional to the index of the subgroup. More precisely, we have the following result. Proposition 10.3.6 Let X be a normal space equipped with a continuous action T : G × X → X of a countable amenable group G. Let H be a subgroup of finite

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10 Mean Topological Dimension for Actions of Amenable Groups

index of G and let T (H ) denote the continuous action of H on X induced by T . Then one has (10.3.2) mdim(X, H, T (H ) ) = [G : H ] mdim(X, G, T ), where [G : H ] denotes the index of H in G. Let us first establish an auxiliary result. Lemma 10.3.7 Let G be a countable amenable group and let H be a subgroup of finite index of G. Suppose that (L n )n≥1 is a Følner sequence for H and let R ⊂ G be a complete set of representatives of the left-cosets of H . Then the sequence (Fn )n≥1 , where Fn := R L n for all n ≥ 1, is a Følner sequence for G. Proof Let k := [G : H ] denote the index of H in G and fix some element g ∈ G. Since G acts on G/H by left-multiplication, there is a permutation σ : R → R and a map ρ : R → H such that gr = σ (r )ρ(r ) for all r ∈ R. Thus, we have that Fn \g Fn =

r L n \g

r ∈R

=

r Ln \

r ∈R

=

=

r Ln \

r Ln

r ∈R

gr L n

σ (r )ρ(r )L n

r ∈R

r Ln \

r ∈R

=

r ∈R

r ∈R

rρ(σ

−1

(r ))L n

r ∈R

r L n \rρ(σ −1 (r ))L n , r ∈R

where denotes disjoint union. Setting h r := ρ(σ −1 (r )) to simplify notation, we deduce that

|Fn \g Fn | =

(r L n \r h r L n )

r ∈R = |r L n \r h r L n | r ∈R

=

r ∈R

|L n \h r L n |.

10.3 General Properties of Mean Topological Dimension

199

As |Fn | = k|L n | and (L n )n≥1 is a Følner sequence for H , we conclude that |Fn \g Fn | 1 |L n \h r L n | = |Fn | k |L n | r ∈R

tends to 0 as n goes to infinity. This shows that (Fn )n≥1 is a Følner sequence for G. Proof of Proposition 10.3.6 Let (L n )n≥1 be a Følner sequence for H and let R ⊂ G be a complete set of representatives of the left cosets of H . We can assume 1G ∈ R. By virtue of Lemma 10.3.7, the sequence (Fn )n≥1 , where Fn := R L n for all n ≥ 1, is a Følner sequence for G. Let α be a finite open cover of X . Let n ≥ 1. As 1G ∈ R, we have that L n ⊂ Fn . By applying Proposition 10.2.1(ii), we deduce that D(α L n ) ≤ D(α Fn ).

(10.3.3)

This gives us D(α L n ) n→∞ |L n | D(α L n ) (since |Fn | = k|L n |) = k lim n→∞ |Fn | D(α Fn ) ≤ k lim (by (10.3.3)) n→∞ |Fn | = k D(α, X, G, T ) ≤ k mdim(X, G, T )

D(α, X, H, T (H ) ) = lim

and hence mdim(X, H, T (H ) ) = sup D(α, X, H, T (H ) ) ≤ k mdim(X, G, T ). α

On the other hand, observe that α Fn = Tg−1 (α) g∈Fn

=

Tg−1 (α)

g∈R L n

=

Tr−1 h (α)

h∈L n ,r ∈R

=

h∈L n ,r ∈R

Th−1 (Tr−1 (α))

(10.3.4)

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10 Mean Topological Dimension for Actions of Amenable Groups

=

Th−1

=

Tr−1 (α)

(by Proposition 6.1.1)

r ∈R

h∈L n

Tr−1 (α)

r ∈R

Ln

so that D(α, X, G, T ) = lim

n→∞

= lim

D(α Fn ) |Fn | −1 D r ∈R Tr (α) L n

|Fn |

n→∞

−1 D r ∈R Tr (α) L n 1 lim = k n→∞ |L n | 1 = D Tr−1 (α), H, T (H ) k r ∈R

1 ≤ mdim(X, H, T (H ) ). k This implies that mdim(X, G, T ) = sup D(α, X, G, T ) ≤ α

1 mdim(X, H, T (H ) ). k

Inequalities (10.3.4) and (10.3.5) give us (10.3.2).

(10.3.5)

Remark 10.3.8 Let f be a homeomorphism of a normal space X and let n be a positive integer. It follows from Proposition 6.4.2 that mdim(X, f n ) = n mdim(X, f ). This equality may also be obtained by applying Proposition 10.3.6 by taking G := Z and H := nZ. Indeed, nZ is a subgroup of Z with index n. On the other hand, if (X, Z, T ) is the dynamical system generated by f , then (X, nZ, T |n Z ) is (canonically topologically conjugate to) the dynamical system generated by f n and we know (cf. Example 10.2.3) that mdim(X, Z, T ) = mdim(X, f ) and mdim(X, nZ, T |n Z ) = mdim(X, f n ). Remark 10.3.9 Although actions of finite groups are uninteresting from a dynamical viewpoint, we can apply Proposition 10.3.6 with G finite and H := {1G }. This shows that if X is a normal space equipped with a continuous action T : G × X → X of a finite group G, then mdim(X, G, T ) =

dim(X ) . |G|

10.4 Metric Approach to Mean Topological Dimension

201

10.4 Metric Approach to Mean Topological Dimension Let X be a compact metrizable space equipped with a continuous action T : G × X → X of a group G. Let us fix some metric d on X that is compatible with the topology. For each g ∈ G, the map Tg : X → X is a homeomorphism of X . Consequently, the map (x, y) → d(Tg (x), Tg (y)) is a metric on X that is also compatible with the topology. More generally, given a non-empty finite subset A ∈ P f in (G) of G, the map d A on X × X defined by d A (x, y) := max d(Tg (x), Tg (y)) for all x, y ∈ X g∈A

is a metric on X compatible with the topology. Recall from Sect. 4.6 (see in particular Proposition 4.6.2) that dimε (X, d A ) is the smallest integer n ≥ 0 such that there exists a compact metrizable space K with dim(K ) = n and a continuous map f : X → K that is ε-injective with respect to the metric d A . If A is the empty set, we will take by convention dimε (X, d A ) := 0. Proposition 10.4.1 Let X be a compact metrizable space equipped with a continuous action T : G × X → X of a group G and let d be a metric on X compatible with the topology. Let ε > 0. Then the following hold: (i) dimε (X, d Ag ) = dimε (X, d A ) for all g ∈ G and A ∈ P f in (G); (ii) dimε (X, d A ) ≤ dimε (X, d B ) for all A, B ∈ P f in (G) such that A ⊂ B; (iii) dimε (X, d A∪B ) ≤ dimε (X, d A ) + dimε (X, d B ) for all A, B ∈ P f in (G). Proof Let g ∈ G and A, B ∈ P f in (G). Assertion (i) immediately follows from the fact that the metric spaces (X, d Ag ) and (X, d A ) are isometric. To see this, observe that the homeomorphism Tg is an isometry from (X, d Ag ) onto (X, d A ) since d A (Tg (x), Tg (y)) = max d(Th (Tg (x)), Th (Tg (y))) h∈A

= max d(Thg (x), Thg (y)) h∈A

= max d(Th (x), Th (y)) h∈Ag

= d Ag (x, y) for all x, y ∈ X . If A ⊂ B, then we have that d A (x, y) ≤ d B (x, y) for all x, y ∈ X and hence dimε (X, d A ) ≤ dimε (X, d B ) by Corollary 4.6.4. This shows (ii). To establish (iii), first observe that d A∪B = max(d A , d B ). Now, let K be a compact metrizable space with dim(K ) = dimε (X, d A ) such that there exists a continuous map f : X → K that is ε-injective for the metric d A . Similarly, let L be a compact metrizable space with dim(L) = dimε (X, d B ) such that there exists a continuous

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10 Mean Topological Dimension for Actions of Amenable Groups

map g : X → L that is ε-injective for the metric d B . Then the map F : X → K × L defined by F(x) = f (x), g(x) is clearly continuous and ε-injective for the metric max(d A , d B ). Since K × L, being the product of two compact metrizable spaces, is itself compact and metrizable, we deduce that dimε (X, d A∪B ) = dimε (X, max(d A , d B )) ≤ dim(K × L). As dim(K × L) ≤ dim(K ) + dim(L) by Corollary 4.5.6, we conclude that dimε (X, d A∪B ) ≤ dim(K ) + dim(L) = dimε (X, d A ) + dimε (X, d B ).

This shows (iii).

Suppose that (X, d) is a compact metric space equipped with a continuous action T : G × X → X of a countable amenable group G. For each ε > 0, we define the real number mdimε (X, d, G, T ) ≥ 0 by mdimε (X, d, G, T ) := lim

n→∞

dimε (X, d Fn ) , |Fn |

where (Fn )n≥1 is a Følner sequence for G. Since the map A → dimε (X, d A ) is right-invariant and subadditive on P f in (G) by assertions (i) and (iii) in Proposition 10.4.1, it follows from Theorem 9.4.1 that the above limit exists, is finite, and does not depend on the choice of the Følner sequence (Fn ). For a fixed n ≥ 1, the map ε → dimε (X, d Fn ) is non-increasing. This implies that the map ε → mdimε (X, d, G, T ) is also non-increasing. We deduce that mdimε (X, d, G, T ) admits a (possibly infinite) limit as ε tends to 0. It turns out that this limit is precisely the mean topological dimension of the dynamical system (X, G, T ). More precisely, we have the following result. Theorem 10.4.2 Let X be a compact metrizable space equipped with a continuous action T : G × X → X of a countable amenable group G. Let d be a metric on X that is compatible with the topology. Then one has mdim(X, G, T ) = lim mdimε (X, d, G, T ). ε→0

(10.4.1)

Proof Consider a finite open cover α of X . Let λ > 0 be a Lebesgue number of α relative to the metric d. We claim that D(α A ) ≤ dimλ (X, d A )

(10.4.2)

for all A ∈ P f in (G). Indeed, suppose that K is a compact metrizable space such that there exists a continuous map f : X → K that is λ-injective with respect to the metric d A . Then, for every y ∈ K and for all g ∈ A, the set Tg ( f −1 (y)) has d-diameter at most λ. As λ is a Lebesgue number for α, this implies that f −1 (y) is

10.4 Metric Approach to Mean Topological Dimension

203

contained in some open set belonging to the cover α A . Hence f is α A -compatible by Lemma 4.5.3. We deduce that D(α A ) ≤ dim(K ) by applying Proposition 4.4.5. This gives us (10.4.2). Consider now a Følner sequence (Fn )n≥1 for G. Using (10.4.2), we get dimλ (X, d Fn ) D(α Fn ) ≤ |Fn | |Fn | for all n ≥ 1. Letting n tend to infinity, we obtain D(α, X, G, T ) ≤ mdimλ (X, d, G, T ). Since, as mentioned above, the map ε → mdimε (X, d, G, T ) is non-increasing, this implies D(α, X, G, T ) ≤ lim mdimε (X, d, G, T ). ε→0

Therefore this yields mdim(X, G, T ) = sup D(α, X, G, T ) ≤ lim mdimε (X, d, G, T ). ε→0

α

To complete the proof, it suffices to prove that, conversely, lim mdimε (X, d, G, T ) ≤ mdim(X, G, T ).

ε→0

(10.4.3)

Let ε > 0. Consider the open cover of X by its open d-balls of radius ε/2. By compactness of X , it admits a finite subcover α. Let A ∈ P f in (G). By Proposition 4.4.6, we can find a polyhedron P such that dim(P) = D(α A ) and a continuous α A compatible map f : X → P. Let y ∈ P. As f is α A -compatible, the set f −1 (y) is contained in one of the open sets of the cover α A . Consequently, for each element g ∈ A, the set Tg ( f −1 (y)) is contained in one of the open balls of radius ε/2. Thus, the map f is ε-injective for the metric d A . As P is compact and metrizable, this implies dimε (X, d A ) ≤ dim(P) = D(α A ). If (Fn )n≥1 is a Følner sequence for G, we deduce that mdimε (X, d, G, T ) = lim

n→∞

dimε (X, d Fn ) D(α Fn ) ≤ lim = D(α, X, G, T ). n→∞ |Fn | |Fn |

Since D(α, X, G, T ) ≤ mdim(X, G, T ), we finally get mdimε (X, d, G, T ) ≤ mdim(X, G, T ), which yields (10.4.3), by letting ε tend to 0.

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10 Mean Topological Dimension for Actions of Amenable Groups

10.5 Shifts and Subshifts Let G be a group. Given g ∈ G, we denote by Rg the right-multiplication by g on G, i.e., the map Rg : G → G defined by Rg (h) := hg for all h ∈ G. Observe that R1G = Id G

and Rg1 g2 = Rg2 ◦ Rg1

for all g1 , g2 ∈ G. Let now K be a topological space. The set K G consists of all maps x : G → K . It may be identified with the product of a family of copies of K indexed by G. We equip K G = g∈G K with the product topology. Consider the map : G × K G → K G defined by (10.5.1)

(g, x) := x ◦ Rg for all g ∈ G and x ∈ K G . Proposition 10.5.1 Let G be a group and K a topological space. Then the map

: G × K G → K G , defined by (10.5.1), is a continuous action of G on K G . Proof For all x ∈ K G and g1 , g2 ∈ G, we have that

(1G , x) = x ◦ R1G = x ◦ Id G = x and

(g1 , (g2 , x)) = (g1 , x ◦ Rg2 ) = x ◦ Rg2 ◦ Rg1 = x ◦ Rg1 g2 = (g1 g2 , x). This shows that is an action of G on K G . To see that this action is continuous, it suffices to observe that if we fix g ∈ G, the element (g, x) is obtained from x ∈ K G by a permutation of its coordinates that is entirely determined by g. Indeed, this shows that g is continuous by definition of the product topology. If G is a group and K a topological space, the continuous action : G × K G → K G of G on K G defined by (10.5.1) is called the G-shift on K G and the dynamical system (K G , G, ) is called the full shift, or simply the shift, with symbol space K over the group G. Note that K G is compact if K is compact (by the Tychonoff product theorem) and that K G is metrizable if K is metrizable and G is countable (since the product of a countable family of metrizable spaces is itself metrizable). Example 10.5.2 If we take G = Z, then (K G , G, ) is the dynamical system generated by the shift map σ : K Z → K Z introduced in Sect. 7.1. A closed G-invariant subset X ⊂ K G is called a subshift of K G . For E ⊂ G, let π E : K G → K E denote the canonical projection map, that is, the map defined by π E (x) := x| E for all x ∈ K G , where x| E denote the restriction of x : G → K to E ⊂ G.

10.5 Shifts and Subshifts

205

The following result provides an upper bound for the mean topological dimension of a subshift. Theorem 10.5.3 Let K be a compact metrizable space of finite topological dimension dim(K ) < ∞. Let G be a countable amenable group and let (Fn )n≥1 be a Følner sequence for G. Suppose that X ⊂ K G is a subshift. Then one has mdim(X, G, ) ≤ lim inf n→∞

dim(π Fn (X )) . |Fn |

(10.5.2)

For the proof, we shall use the following general properties of Følner sequences. Lemma 10.5.4 Let G be a countable amenable group and let (Fn )n≥1 be a Følner sequence for G. Let S be a finite subset of G. Then one has |S Fn \Fn | = 0. n→∞ |Fn | lim

(10.5.3)

Proof We have that

|S Fn \Fn | =

(g Fn \Fn )

g∈S

≤ |g Fn \Fn | g∈S

=

|Fn \g Fn |

(since |g Fn | = |Fn | for all g ∈ S).

g∈S

After dividing by |Fn | and letting n tend to infinity, this gives us (10.5.3) since (Fn )n≥1 is a Følner sequence for G. Proof Let us choose a metric d on K G which is compatible with the topology. Let ε > 0. By compactness of K G , we can find a finite subset S ⊂ G such that, for all x, y ∈ K G , (10.5.4) π S (x) = π S (y) ⇒ d(x, y) < ε. By replacing S by S ∪ {1G }, we can assume 1G ∈ S. Let A be a non-empty finite subset of G. Let f : X → K S A denote the restriction of π S A to X . If x, y ∈ X satisfy f (x) = f (y), then we have that π S (gx) = π S (g y) for all g ∈ A. This implies d A (x, y) < ε by (10.5.4). Consequently, the map f is ε-injective with respect to the metric d A . As f (X ) ⊂ K S A is a compact metrizable space, we deduce that dimε (X, d A ) ≤ dim( f (X )).

(10.5.5)

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10 Mean Topological Dimension for Actions of Amenable Groups

Observe now that A ⊂ S A since 1G ∈ S. Thus, we have the inclusion f (X ) ⊂ π A (X ) × K S A\A . It follows that dim( f (X )) ≤ dim(π A (X ) × K S A\A ) ≤ dim(π A (X )) + dim(K

(by Proposition 1.2.1) S A\A

)

≤ dim(π A (X )) + |S A\A| dim(K )

(by Corollary 4.5.6) (again by Corollary 4.5.6).

After dividing by |A|, replacing A by Fn , and using (10.5.5), we deduce that dimε (X, d Fn ) dim(π Fn (X )) |S Fn \Fn | ≤ + dim(K ) |Fn | |Fn | |Fn | for all n ≥ 1. By letting n tend to infinity and using the result of Lemma 10.5.4, we obtain dim(π Fn (X )) . mdimε (X, d, G, ) ≤ lim inf n→∞ |Fn | Finally, by letting ε tend to 0 and using the result of Theorem 10.4.2, this gives us (10.5.2). As a consequence of Theorem 10.5.3, we get the following extension of Theorem 7.1.3. Corollary 10.5.5 Let K be a compact metrizable space and let G be a countablyinfinite amenable group. Then one has mdim(K G , G, ) ≤ stabdim(K ) ≤ dim(K ).

(10.5.6)

Proof If we take X = K G , then π Fn (X ) is homeomorphic to K |Fn | . As limn→∞ |Fn | = ∞ by Lemma 10.3.3, Inequality (10.5.6) immediately follows from (10.5.2) and the definition of stabdim(X ) given in Sect. 6.2.

10.6 Mean Topological Dimension of Shifts over Polyhedra The following result is an extension of Theorem 7.2.1. Theorem 10.6.1 Let N ∈ N and let K := [0, 1] N be the N -dimensional cube. Let G be a countably-infinite amenable group. Then one has mdim(K G , G, ) = N .

10.6 Mean Topological Dimension of Shifts over Polyhedra

207

Proof We immediately get mdim(K G , G, ) ≤ dim(K ) = N by applying Corollary 10.5.5 and Theorem 3.5.4. Thus, it only remains to prove that mdim(K G , G, ) ≥ N . Let d K be the metric on K associated with the sup-norm · ∞ on R N . Since G is countably-infinite, we can find a family of positive real numbers (αg )g∈G such that α1G = 1 and

αg < ∞

g∈G

and consider the metric d on K G defined by d(x, y) :=

αg d K (x(g), y(g))

g∈G

for all x, y ∈ K G . Clearly d is compatible with the topology of K G . Moreover, we have that αg d K (x(g), y(g)) = d(x, y) d K (x(1G ), y(1G )) = α1G d K (x(1G ), y(1G )) ≤ g∈G

(10.6.1) for all x, y ∈ K G . Let A be a non-empty finite subset of G. Consider the metric d A on K G defined by d A (x, y) := max d g (x), g (y) . g∈A

Inequality (10.6.1) gives us max d K (x(g), y(g)) = max d K ( g (x)(1G ), g (y)(1G )) ≤ max d g (x), g (y) g∈A

g∈A

g∈A

= d A (x, y)

(10.6.2)

for all x, y ∈ K G . Consider now the topological embedding ϕ : K A → K G that sends each u ∈ K A to the element x = ϕ(u) ∈ K G defined by x(g) :=

u(g) 0

if g ∈ A, otherwise,

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10 Mean Topological Dimension for Actions of Amenable Groups

and the metric ρ on K A defined by ρ(u, v) = max d K (u(g), v(g)) g∈A

for all u, v ∈ K A . Inequality (10.6.2) implies that ρ(u, v) ≤ d A ϕ(u), ϕ(v) , for all u, v ∈ K A . By applying Proposition 4.6.3, we deduce that dimε (K A , ρ) ≤ dimε (K G , d A )

(10.6.3)

for all ε > 0. Observe now that the metric space (K A , ρ) is isometric to ([0, 1] N |A| , ρ ), where ρ is the metric associated with the sup-norm on R N |A| . Therefore, it follows from Proposition 4.6.5 that dimε (K A , ρ) = N |A| for all 0 < ε ≤ 1. Thus, Inequality (10.6.3) gives us N |A| ≤ dimε (K G , d A ) for all 0 < ε ≤ 1. After replacing A by Fn , where (Fn )n≥1 is a Følner sequence for G, we deduce that dimε (K G , d Fn ) ≥N n→∞ |Fn |

mdimε (K G , d, G, ) = lim

for all 0 < ε ≤ 1. By applying Theorem 10.4.2, we conclude that mdim(K G , G, ) = lim mdimε (K G , d, G, ) ≥ N . ε→0

This shows (10.6).

Corollary 10.6.2 Let N ∈ N and let K be a metrizable or compact Hausdorff space such that there exists a subset A ⊂ K that is homeomorphic to the N -cube [0, 1] N . Let G be a countably-infinite amenable group. Then one has mdim(K G , G, ) ≥ N . Proof The subset A G ⊂ K G is closed and -invariant. We deduce that mdim(K G , G, ) ≥ mdim(A G , G, ) = N by applying Proposition 10.3.4. Corollary 10.6.3 Let K be a polyhedron. Let G be a countably-infinite amenable group. Then one has mdim(K G , G, ) = dim(K ). Proof Let N := dim(K ). Since every polyhedron is compact and metrizable, we have that mdim(K G , G, ) ≤ N by Corollary 10.5.5. On the other hand, we know that K contains a subset A homeomorphic to an N -simplex and hence to [0, 1] N . We deduce that mdim(K G , G, ) ≥ N = dim(K ) by applying Corollary 10.6.2.

10.6 Mean Topological Dimension of Shifts over Polyhedra

209

Corollary 10.6.4 Let K = [0, 1]N be the Hilbert cube. Let G be a countably-infinite amenable group. Then one has mdim(K G , G, ) = ∞. Proof Let N be a positive integer. The subset A ⊂ K consisting of all (u n )n∈N ∈ K such that u n = 0 for all n ≥ N is homeomorphic to [0, 1] N . Thus, we deduce from Corollary 10.6.2 that mdim(K G , G, ) ≥ N .

10.7 Subshifts of Block-Type In this section, we assume that K is a compact metrizable space. Let G be a group. Suppose that H is a subgroup of finite index of G and let R ⊂ G be a complete set of representatives of the left-cosets of H . Thus, R is a finite set with |R| = [G : H ] and every element g ∈ G can be uniquely written in the form g = r h with r ∈ R and h ∈ H . Let B be a closed subset of K R . We define the subset X 0 = X 0 (K , G, H, R, B) ⊂ K G by X 0 := {x ∈ K G | π R ( h (x)) ∈ B for all h ∈ H }

(10.7.1)

and the subset X = X (K , G, H, R, B) ⊂ K G by X :=

g (X 0 ).

(10.7.2)

g∈G

Proposition 10.7.1 The subsets X 0 , X ⊂ K G defined above satisfy the following properties: (i) X 0 ⊂ X ; (ii) X 0 is a closed (H ) -invariant subset of K G ; (iii) the dynamical system (X 0 , H, (H ) ) is topologically conjugate to the H -shift on B H; (iv) X = r ∈R r (X 0 ); (v) the subset X ⊂ K G is a subshift (i.e., a closed -invariant subset of K G ). Proof Property (i) follows from the fact that X 0 = 1G (X 0 ) ⊂ X. The definition of X 0 may be written in the form X0 =

h∈H

(π R ◦ h )−1 (B).

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10 Mean Topological Dimension for Actions of Amenable Groups

As π R ◦ h is continuous for all h ∈ H and B is closed in K R , this shows that X 0 is the intersection of a family of closed subsets of K G and hence closed in K G . Moreover, we clearly have k (x) ∈ X 0 for all x ∈ X 0 and k ∈ H , so that X 0 is

(H ) -invariant. Consequently, X 0 satisfies (ii). Consider the map ϕ : X 0 → B H defined by ϕ(x)(h) := π R ( h (x)) for all x ∈ X 0 and h ∈ H . Note that ϕ is well defined by (10.7.1) and that ϕ is bijective since the family of cosets (Rh)h∈H is a partition of G. As X 0 is compact and ϕ is clearly continuous, we deduce that ϕ is a homeomorphism from X 0 onto B H . On the other hand, denoting by the H -shift on B H , we have, for all x ∈ X 0 and h, k ∈ H , (ϕ ◦ h )(x)(k) = ϕ( h (x))(k) = π R ( k ( h (x))) = π R ( kh (x)) = ϕ(x)(kh) = h (ϕ(x))(k) = ( h ◦ ϕ)(x)(k). It follows that (ϕ ◦ h )(x) = ( h ◦ ϕ)(x) for all x ∈ X 0 . This implies that ϕ ◦ h =

h ◦ ϕ for all h ∈ H . Thus, ϕ is a topological conjugacy between the dynamical systems (X 0 , H, (H ) ) and (B H , H, ). This shows (iii). Property (iv) follows from the fact that X=

g (X 0 )

g∈G

=

r h (X 0 )

r ∈R,h∈H

=

r ◦ h (X 0 )

r ∈R,h∈H

=

r ( h (X 0 ))

r ∈R,h∈H

=

r (X 0 )

(since X 0 is (H ) -invariant by (ii)).

r ∈R

Since X 0 is closed in K G , we deduce from (iv) that X is a finite union of closed subsets of K G and hence closed in K G . As X ⊂ K G is clearly -invariant by (10.7.2), we conclude that X ⊂ K G is a subshift.

10.7 Subshifts of Block-Type

211

One says that the subshift X ⊂ K G defined by (10.7.2) is the subshift of block-type associated with the quintuple (K , G, H, R, B).

10.8 Construction of Subshifts with Prescribed Mean Dimension It follows from Corollary 10.6.3 and Proposition 10.3.4 that if G is a countable amenable group and P is a polyhedron, then every subshift X ⊂ P G satisfies mdim(X, G, ) ≤ dim(P). Conversely, it is natural to ask whether every real number in the interval [0, dim(P)] appears as the mean topological dimension of some subshift X ⊂ P G . The following result gives a partial answer to this question. Theorem 10.8.1 Let G be a countable amenable group and let P be a polyhedron. Suppose that G contains subgroups of arbitrarily large finite index (i.e., for any integer N ≥ 1, there exists a subgroup H of G such that N ≤ [G : H ] < ∞). Then, for every real number ρ satisfying 0 ≤ ρ ≤ dim(P), there exists a subshift X ⊂ P G such that mdim(X, G, ) = ρ. For the proof, we shall use several lemmas. Lemma 10.8.2 Let K be a compact metrizable space with dim(K ) < ∞. Let G be a countably-infinite amenable group. Suppose that H is a subgroup of finite index of G. Let R ⊂ G be a complete set of representatives of the left-cosets of H in G and let B ⊂ K R be a polyhedron. Let X ⊂ K G be the subshift of block type associated with the quintuple (K , G, H, R, B). Then one has mdim(X, G, ) ≥

dim(B) . [G : H ]

(10.8.1)

Proof Let X 0 ⊂ K G as in (10.7.1). It follows from Proposition 10.3.6 that mdim(X, G, ) =

mdim(X, H, (H ) ) . [G : H ]

On the other hand, as X 0 is a (H ) -invariant subspace of X and (X 0 , H, (H ) ) is topologically conjugate to (B H , H, ), where denotes the H -shift on B H , by Proposition 10.7.1, we have that mdim(X, H, (H ) ) ≥ mdim(X 0 , H, (H ) )

(by Proposition 10.3.1)

= mdim(B , H, )

(by Proposition 10.3.1)

= dim(B)

(by Corollary 10.6.3).

H

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10 Mean Topological Dimension for Actions of Amenable Groups

Thus, we get mdim(X, G, ) ≥

dim(B) . [G : H ]

Lemma 10.8.3 Let K be a compact metrizable space with dim(K ) < ∞. Let G be a countably-infinite amenable group. Suppose that H is a normal subgroup of finite index of G. Let R ⊂ G be a complete set of representatives of the cosets of H in G and let (Br )r ∈R be a family of closed subsets of K . Let B denote the closed subset of K R defined by B := r ∈R Br and let X ⊂ K G be the subshift of block type associated with the quintuple (K , G, H, R, B). Then one has mdim(X, G, ) ≤

dim(B) . [G : H ]

(10.8.2)

Proof Let h ∈ H and g ∈ G. We claim that dim(π Rh ( g (X 0 ))) ≤ dim(B).

(10.8.3)

Indeed, first observe that π Rh ( g (X 0 )) ⊂

π{r h} ( g (X 0 )).

r ∈R

Now, using the canonical identification of K {r h} with K , we have that π{r h} ( g (X 0 )) = { g (x)(r h) | x ∈ X 0 } = {x(r hg) | x ∈ X 0 } for all r ∈ R. Since H is assumed to be normal in G, there is a bijective map σ : R → R and a map ρ : R × H → H (both depending on g) such that r hg = σ (r )ρ(r, h). This gives us π{r h} ( g (X 0 )) = {x(σ (r )ρ(r, h)) | x ∈ X 0 } = { ρ(r,h) (x)(σ (r )) | x ∈ X 0 } = {x(σ (r )) | x ∈ ρ(r,h) (X 0 )} = {x(σ (r )) | x ∈ X 0 } (since X 0 is (H ) -invariant by Proposition 10.7.1) = Bσ (r ) . We deduce that π Rh ( g (X 0 )) is a closed subset of r ∈R Bσ (r ) . As r ∈R Bσ (r ) is homeomorphic to r ∈R Br = B, this shows (10.8.3).

10.8 Construction of Subshifts with Prescribed Mean Dimension

213

Now, let L be a non-empty finite subset of H . By Proposition 10.7.1, we have that X=

g (X 0 ).

g∈R

This implies that π R L (X ) = π R L (

g (X 0 )) =

g∈R

π R L ( g (X 0 )),

g∈R

and hence, by using Corollary 1.2.6, dim(π R L (X )) = max dim(π R L ( g (X 0 ))). g∈R

As π R L ( g (X 0 )) ⊂

π Rh ( g (X 0 )),

h∈L

this gives us dim(π R L (X )) ≤

dim(π Rh ( g (X 0 ))).

h∈L

Applying Inequality (10.8.3), we finally get dim(π R L (X )) ≤ |L| dim(B).

(10.8.4)

Suppose now that (L n )n≥1 is a Følner sequence for H . Then the sequence (Fn )n≥1 , where Fn := R L n for all n ≥ 1, is a Følner sequence for G by Lemma 10.3.7. After replacing L by L n in (10.8.4), we get dim(π Fn (X )) = dim(π R L n (X )) ≤ |L n | dim(B) =

|Fn | dim(B) [G : H ]

for all n ≥ 1. As mdim(X, G, ) ≤ lim inf n→∞

by Theorem 10.5.3, this yields (10.8.2).

dim(π Fn (X )) |Fn |

Lemma 10.8.4 Let G be a countably-infinite amenable group. Suppose that H is a normal subgroup of finite index of G and let R ⊂ G be a complete set of representatives of the cosets of H in G. Let P be a non-empty polyhedron and let p0 ∈ P. Let S ⊂ R and consider the closed subset B ⊂ P R defined by

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10 Mean Topological Dimension for Actions of Amenable Groups

B := P S × { p0 } R\S ⊂ P S × P R\S = P R . Then the subshift of block type X ⊂ B G associated with the quintuple (P, G, H, R, B) satisfies |S| mdim(X, G, ) = dim(P). [G : H ] Proof This immediately follows from Lemmas 10.8.2 and 10.8.3 since dim(B) = |S| dim(P) by Corollary 3.5.11. Lemma 10.8.5 Let G be a group containing subgroups of arbitrarily large finite index. Then there exists a strictly decreasing sequence (Hn )n∈N of normal subgroups of finite index of G. Proof We use induction. We start by taking H0 = G. Suppose that H0 , H1 , . . . , Hn are normal subgroups of finite index of G such that H0 H1 · · · Hn . Since G contains subgroups of arbitrarily large finite index, we can find a subgroup of finite index K ⊂ G such that [G : K ] > [G : Hn ]. Then K ∩ Hn is a subgroup of finite index of G which is strictly contained in Hn . Moreover, it follows from Lemma 9.2.23 that we can find a normal subgroup of finite index L of G such that L ⊂ K ∩ Hn . As Hn K ∩ Hn ⊃ L , we can take Hn+1 = L.

Proof of Theorem 10.8.1 Let λ := ρ/ dim(P). If ρ = dim(P), we can take X = P G by Corollary 10.6.3. Thus, we can assume 0 ≤ λ < 1. By applying Lemma 10.8.5, we can find a strictly decreasing sequence (Hn )n∈N of normal subgroups of finite index of G with H0 = G. Let νn := [G : Hn ] denote the index of Hn . Observe that ν0 = 1 and that νn+1 = [Hn : Hn+1 ] ≥ 2 νn for all n ∈ N. Let an denote the integral part of νn λ and bn := an + 1. We then have a0 = 0, b0 = 1, and (10.8.5) an ≤ νn λ < bn for all n ∈ N. It follows that the sequences (u n ) and (vn ) defined by u n :=

an bn and vn := νn νn

10.8 Construction of Subshifts with Prescribed Mean Dimension

215

satisfy u n ≤ λ < vn

(10.8.6)

and vn − u n =

bn − an 1 = νn νn

(10.8.7)

for all n ∈ N. As νn → ∞ as n goes to infinity, we deduce from (10.8.6) and (10.8.7) that the sequences (u n ) and (vn ) both converge to λ. Observe also that (10.8.5) implies that νn+1 νn+1 an ≤ νn+1 λ < bn νn νn so that νn+1 νn+1 an ≤ an+1 < bn+1 ≤ bn νn νn

(10.8.8)

for all n ∈ N. Let us choose, for each n ∈ N, a complete set Rn ⊂ G of representatives for the cosets of Hn in G. Let μn : Rn+1 → Rn denote the map which associates to each element r ∈ Rn+1 the representative of the coset r Hn . Note that μn is surjective and that every element of Rn is the image by μn of exactly νn+1 νn elements of Rn+1 . Let us show that we can find two sequences of sets (In )n∈N and (Jn )n∈N with the following properties: (P1) (P2) (P3) (P4) (P5)

In ⊂ Jn ⊂ Rn , |In | = an , |Jn | = bn , μ−1 n (In ) ⊂ In+1 , Jn+1 ⊂ μ−1 n (Jn ),

for all n ∈ N. We proceed by induction on n. We start by taking I0 = ∅ and J0 = R0 . Now, suppose that Ik and Jk have already been constructed for k ≤ n with the required properties. As In ⊂ Jn by (P1), we have that −1 μ−1 n (In ) ⊂ μn (Jn ).

On the other hand, since |In | = an and |Jn | = bn by (P2) and (P3), the fact that μn is νn+1 νn -to-one implies that |μ−1 n (In )| =

νn+1 νn+1 an and |μ−1 bn . n (Jn )| = νn νn

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10 Mean Topological Dimension for Actions of Amenable Groups

Thus, we deduce from (10.8.8) that we can find subsets In+1 and Jn+1 of Rn+1 such that −1 μ−1 n (In ) ⊂ In+1 ⊂ Jn+1 ⊂ μn (Jn ) with |In+1 | = an+1 and |Jn+1 | = bn+1 . This completes our induction. Now, let us fix some point p0 ∈ P and consider the subsets An ⊂ Bn ⊂ P Rn defined by An := P In × { p0 } Rn \In and Bn := P Jn × { p0 } Rn \Jn . Let Y (n) ⊂ P G and Z (n) ⊂ P G denote the subshifts of block-type respectively associated with (P, G, Hn , Rn , An ) and (P, G, Hn , Rn , Bn ). This means that Y (n) =

g∈G

(n)

g (Y0 ) and Z (n) =

(n)

g (Z 0 ),

g∈G

where Y0(n) := {x ∈ P G | π Rn ( h (x)) ∈ An for all h ∈ Hn } / In Hn }, = {x ∈ P G | x(g) = p0 if g ∈ and (n)

Z 0 := {x ∈ P G | π Rn ( h (x)) ∈ Bn for all h ∈ Hn } / Jn Hn }, = {x ∈ P G | x(g) = p0 if g ∈ (we recall that π Rn : P G → P Rn denotes the canonical projection map). Let us show that (Q1) Y (n) ⊂ Z (n) , (Q2) Y (n) ⊂ Y (n+1) , (Q3) Z (n+1) ⊂ Z (n) for all n ∈ N. Property (Q1) follows from the inclusion In ⊂ Jn since it clearly implies Y0(n) ⊂ (n) Z0 . Consider an element g ∈ In Hn . Then g = r h, where r ∈ In and h ∈ Hn . Write g = r h , where r ∈ Rn+1 and h ∈ Hn+1 . Observe that μn (r ) = r since h −1 h ∈ Hn . This implies r ∈ In+1 by (P4). Thus In Hn ⊂ In+1 Hn+1 . This clearly (n) (n+1) implies Y0 ⊂ Y0 and hence (Q2). Similarly, we deduce Jn+1 Hn+1 ⊂ Jn Hn from (P5). This implies Z 0(n+1) ⊂ Z 0(n) , which gives (Q3).

10.8 Construction of Subshifts with Prescribed Mean Dimension

217

Consider now the subshift X ⊂ P G defined by X :=

Z (n) .

n∈N

By using Lemma 10.8.4, we deduce from (P2) and (P3) that mdim(Y (n) , G, ) =

|In | an dim(P) = dim(P) = u n dim(P) νn νn

mdim(Z (n) , G, ) =

|Jn | bn dim(P) = dim(P) = vn dim(P). νn νn

and

Since X ⊂ Z (n) , we get mdim(X, G, ) ≤ mdim(Z (n) , G, ) = vn dim(P) for all n ∈ N by applying Proposition 10.3.4. Letting n tend to infinity, this gives us mdim(X, G, ) ≤ λ dim(P) = ρ.

(10.8.9)

On the other hand, since Y (n) ⊂ Y (N ) ⊂ Z (N ) for all N ≥ n by (Q1) and (Q2), we get Z (N ) = X. Y (n) ⊂ N ≥n

Applying again Proposition 10.3.4, we obtain u n dim(P) = mdim(Y (n) , G, ) ≤ mdim(X, G, ) for all n ∈ N. Letting n tend to infinity, this yields ρ = λ dim(P) ≤ mdim(X, G, ). Inequalities (10.8.9) and (10.8.10) imply that mdim(X, G, ) = ρ.

(10.8.10)

Remark 10.8.6 The group Z clearly satisfies the hypotheses of Theorem 10.8.1, so that Theorem 10.8.1 is an extension of Corollary 7.6.4. More generally, every countable amenable group G admitting a surjective homomorphism f : G → Z satisfies the hypotheses of Theorem 10.8.1 since f −1 (nZ) is then a subgroup of index n of G for every n ≥ 1. This shows in particular that any infinite finitely generated abelian group (e.g., G = Zd for d ≥ 1) satisfies the hypotheses of Theorem 10.8.1.

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10 Mean Topological Dimension for Actions of Amenable Groups

Notes Mean topological dimension for actions of amenable groups was introduced by Gromov in [44]. Its properties were investigated in depth for Z-actions by Lindenstrauss and Weiss in [74]. The exposition in the present chapter closely follows that in [24]. One can define mean topological dimension for actions of uncountable amenable groups by replacing Følner sequences by Følner nets (see the Notes on Chap. 9). The notion of mean topological dimension has been extended to continuous actions of countable sofic groups by Li [68]. Sofic groups were introduced by Gromov [43] and Weiss [115]. The class of sofic groups is a very vast one. It is known to include in particular all residually finite groups and all amenable groups. Actually, the question of the existence of a non-sofic group is still open. For an introduction to the theory of sofic groups, the reader is referred to the survey paper [87] and to [22, Chap. 7]. Theorem 10.8.1 was obtained by Krieger and the author in [24]. Every residually finite countably-infinite amenable group, and hence every infinite finitely generated linear group (see the Notes on Chap. 9), satisfies the hypotheses of Theorem 10.8.1 (see Exercise 10.8). However, there exist countably-infinite amenable groups, such as the infinite finitely generated amenable simple groups exhibited in [52], that do not satisfy the hypotheses of Theorem 10.8.1. It might be interesting to know whether the conclusion of this theorem remains valid for such groups. There is an impressive literature dealing with shifts and subshifts over G = Zd (see for example the survey papers [69, 71] as well as the references therein). For d ≥ 2, the study of subshifts of finite type over Zd has connections with undecidability questions for tilings of Euclidean spaces. In his Ph.D. thesis [51, Corollary 4.2.1], Jaworski proved that if G is an abelian group and X is a compact metrizable space with dim(X ) < ∞, then every minimal dynamical system (X, G, T ) can be embedded in the G-shift on RG (see Exercise 10.8). On the other hand, Krieger [62] has shown that if P is a polyhedron and G is a countably-infinite amenable group, then there exist minimal subshifts X ⊂ P G whose mean topological dimension is arbitrarily close to dim(P). It follows in particular from Krieger’s result that there exist minimal dynamical systems (X, G, T ), where X is compact and metrizable, that do not embed in the G-shift on RG .

Exercises 10.1 Let G be a group and K a topological space. The set K G = g∈G K is equipped with the product topology. For g ∈ G, denote by L g the leftmultiplication by g, i.e., the map L g : G → G defined by L g (h) = gh for all : G × K G → K G defined by h ∈ G. Consider the map (x) = x ◦ L g−1

Exercises

219

is a continuous action of G on K G for all g ∈ G and x ∈ K G . Show that G ) are topologically and that the dynamical systems (K , G, ) and (K G , G, conjugate. Hint: use the map f : K G → K G defined by f (x)(g) := x(g −1 ) for all x ∈ K G and g ∈ G. 10.2 Let X be a compact metrizable space equipped with a continuous action T : G × X → X of a group G. Let S(X ) denote the set consisting of all continuous maps f : X → X . We equip S(X ) with the topology of uniform convergence (i.e., the topology associated with the metric ρ given by ρ( f 1 , f 2 ) := supx∈X d( f 1 (x), f 2 (x)) for all f 1 , f 2 ∈ S(X ), where d is a metric on X that is compatible with the topology). Consider the map U : G × S(X ) → S(X ) given by U (g, f ) := T (g, f (x)) for all g ∈ G, f ∈ S(x), and x ∈ X . (a) Show that U is a continuous action of G on S(X ). (b) Show that the system (X, G, T ) embeds in the system (S(X ), G, U ). Hint: consider the map ι : X → S(x) that sends each a ∈ X to the constant map ι(a) ∈ S(X ) defined by ι(a)(x) := a for all x ∈ X . 10.3 (cf. Exercise 4.11). Let (X, d) be a compact metric space equipped with a continuous action T : G × X → X of a countable amenable group G. We associate to each non-empty finite subset A ⊂ G the metric d A on X defined by d A (x, y) := maxg∈A d(Tg (x), Tg (y)). (a) Let ε > 0 and let (Fn )n≥1 be a Følner sequence for G. Show that the limit mWidimε (X, d, G, T ) := lim

n→∞

Widimε (X, d Fn ) |Fn |

exists, is finite, and does not depend on the choice of the Følner sequence (Fn )n≥1 . (b) Show that mdim(X, G, T ) = lim mWidimε (X, d, G, T ). ε→0

10.4 (cf. Exercise 7.2). Let G be a group and let X be a topological space. One says that a continuous action T of G on X is topologically mixing if, given any pair U, V ⊂ X of non-empty open subsets of X , the set of g ∈ G such that Tg (U ) ∩ V = ∅ is finite. Let K be a topological space. Show that the G-shift on K G is topologically mixing. 10.5 Let G be a group. Let K and L be topological spaces. Show that the G-shift on L G embeds in the G-shift on K G if and only if the space L embeds in K . 10.6 Let X be a non-empty topological space equipped with a continuous action T : G × X → X of a countable amenable group G.

220

10 Mean Topological Dimension for Actions of Amenable Groups

(a) Let α be a finite open cover of X . Show that the map h : P f in (G) → R defined by h(A) = log N (α A ) (where α A is defined by (10.2.1) and N (·) is defined in Exercise 6.11) is right-invariant and subadditive. (b) Let (Fn )n≥1 be a Følner sequence for G. Show that the limit h top (α, X, G, T ) := lim

n→∞

log N (α Fn ) , |Fn |

exists, is finite, and does not depend on the choice of the Følner sequence (Fn ). The quantity 0 ≤ h top (X, G, T ) ≤ ∞ defined by h top (X, G, T ) := sup h top (α, X, G, T ), α

where α runs over all finite open covers of X , is called the topological entropy of the dynamical system (X, G, T ). (c) Let Y be a topological space equipped with a continuous action S : G × Y → Y of G. Suppose that there exists a surjective continuous map f : Y → X such that f ◦ Sg = Tg ◦ f for all g ∈ G. Show that one has h top (X, G, T ) ≤ h top (Y, G, S). (d) Let ϕ : X → X be a homeomorphism of X . Show that the dynamical system (X, Z, T ) generated by ϕ satisfies h top (X, Z, T ) = h top (X, ϕ), where h top (X, ϕ) is the topological entropy of (X, ϕ) (cf. Exercise 6.11). 10.7 Let K be a finite discrete topological space with cardinality k and let G be a countable amenable group. Given A ∈ P f in (G), let π A : K G → K A denote the restriction map. Let X ⊂ K G be a non-empty subshift. (a) Let (Fn )n≥1 be a Følner sequence for G. Show that the limit log |π Fn (X )| n→∞ |Fn |

h(X ) := lim

exists, is finite, does not depend on the choice of the Følner sequence (Fn ), and satisfies 0 ≤ h(X ) ≤ log k. This limit is called the entropy of the subshift X . (b) Show that h(X ) = h top (X, G, ), where denotes the G-shift on X and h top (X, G, ) is the topological entropy of the dynamical system (X, G, ) (cf. Exercise 10.8). (c) Show that if Y ⊂ K G is a subshift such that X ⊂ Y , then h(X ) ≤ h(Y ). 10.8 One says that a group G is residually finite if the intersection of all the finite index subgroups of G is reduced to the identity element (cf. Exercise 2.5). (a) Let G be a group. Show that the following conditions are equivalent: (1) G is residually finite; (2) the intersection of all the finite index normal

Exercises

(b) (c) (d) (e)

(f)

(g)

221

subgroups of G is reduced to the identity element; (3) for every element g = 1G in G, there exist a finite group F and a homomorphism φ : G → F such that φ(g) = 1 F ; (4) for any two distinct elements g1 , g2 ∈ G, there exist a finite group F and a homomorphism φ : G → F such that φ(g1 ) = φ(g2 ); (5) for every finite subset ⊂ G, there exist a finite group F and a homomorphism φ : G → F whose restriction to is injective. Show that every finite group is residually finite. Show that every finitely generated abelian group is residually finite. Show that the additive group Q of rational numbers is not residually finite. Prove that if G is an infinite residually finite group, then G contains subgroups of arbitrarily large finite index. (This shows in particular that every countably-infinite residually finite amenable group satisfies the hypotheses of Theorem 10.8.1.) Show that if G is a residually finite group and K is a topological space, then the periodic points in K G (i.e., the points whose orbit under the G-shift is finite) are dense in K G . Let G be a group. Show that if there exists a Hausdorff topological space K with more than one point such that the periodic points are dense in K G , then G is residually finite.

10.9 Show that every infinite, finitely generated, virtually solvable group satisfies the hypotheses of Theorem 10.8.1. Hint: prove that every infinite, finitely generated, solvable group G has subgroups of arbitrarily large finite index by induction on the solvability degree of G. 10.10 Let X be a topological space equipped with a continuous action T : G × X → X of a group G. Suppose that K is a topological space such that X embeds in K . Show that the dynamical system (X, G, T ) embeds in the G-shift on K G . 10.11 Let X be a compact metrizable space equipped with a continuous action T : G × X → X of a group G. Suppose that dim(X ) < ∞. Show that there exists an integer n ≥ 1 such that the dynamical system (X, G, T ) embeds in the shift ((Rn )G , G, ). 10.12 Let X be a compact space equipped with a continuous action T : G × X → X of a group G. Let K be a Hausdorff space. Show that the following conditions are equivalent: (1) the dynamical system (X, G, T ) embeds in the shift (K G , G, ); (2) there exists a continuous map f : X → K such that, given any two distinct points x and y in X , there is an element g ∈ G satisfying f (Tg (x)) = f (Tg (y)). 10.13 Let G be a group. One says that an action T : G × X → X of G on a set X is free if Tg (x) = x for all g ∈ G\{1G } and x ∈ X . Show that if K is a topological space having more than one point, then the G-shift on K G is a free action. 10.14 (An embedding theorem for free actions [51, Theorem 4.2]). Let G be an infinite group and let X be a compact metrizable space equipped with a continuous action T : G × X → X of G. We suppose that the action of G on X is free (cf. Exercise 10.13) and that X has finite topological dimension dim(X ) < ∞.

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10 Mean Topological Dimension for Actions of Amenable Groups

(a) Let x and y be distinct points in X . Show that for every integer m ≥ 1, there exist elements g1 , . . . , gm ∈ G such that the points Tg1 (x), . . . , Tgm (x), Tg1 (y), . . . , Tgm (y) are pairwise distinct. (b) Show that (X, G, T ) embeds in the G-shift on RG by using the result of Exercise 10.12 and following the lines of the proof of Theorem 8.3.1. 10.15 (An embedding theorem for minimal actions of abelian groups [51, Corollary 4.2.1]). Let G be an abelian group and let X be a compact metrizable space equipped with a continuous action T : G × X → X of G. We suppose that the action of G on X is minimal (i.e., every orbit is dense in X ) and that X has finite topological dimension dim(X ) < ∞. (a) Let x ∈ X . Show that H := {g ∈ G| Tg (x) = x} is a subgroup of G and that H does not depend on the choice of the point x ∈ X . (b) Suppose that the group G/H is finite. Show that X is finite and then deduce from the result of Exercise 10.10 that the dynamical system (X, G, T ) embeds in the G-shift on RG . (c) Suppose now that the quotient group Q := G/H is infinite. Observe that T induces a continuous free action of Q on X . Then conclude that (X, G, T ) embeds in the G-shift on RG by using the result of Exercise 10.14 and the canonical embedding R Q → RG .

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Index

A Abstract simplicial complex, 64 Accessible space, 9 Action continuous group —, 192 free —, 221 group —, 191 topologically mixing —, 219 Adding machine, 154 Affine — dimension, 49 — hyperplane, 49 Affine subspace, 49 direction of an —, 49 Affinely independent points, 50 Almost-periodic, 154 α-compatible map, 75 Amenable group, 158, 161

B Barycenter of a simplex, 54 Barycentric subdivision, 57 N -th —, 59 Base — of a free group, 169 — of a topological space, 31 neighborhood —, 31 Baumslag-Solitar group, 166 Bing space, 94 Block-type subshift of —, 131, 211 ˇ Brouwer-Cech dimension, 21

C Cantor

— space, 28 — teepee, 102 — ternary set, 28 Cellular automaton, 137 Chain of simplices, 54 length of a —, 54 Clopen subset, 31 Closed cover, 4 Cofinite topology, 6 Combinatorially equivalent families of sets, 14 Commutator, 167 Completely normal space, 22 Component connected —, 40 Concave map, 120 Connected — component, 40 — space, 9 Coset left- —, 197 Cover, 4 closed —, 4 open —, 4

D Defining law, 129 Derived — series, 167 — subgroup, 167 Diameter, 10 Dimension — mean topological, 194 — of a topological space, 5 affine —, 49 ˇ Brouwer-Cech —, 21

© Springer International Publishing Switzerland 2015 M. Coornaert, Topological Dimension and Dynamical Systems, Universitext, DOI 10.1007/978-3-319-19794-4

229

230 ˇ Cech-Lebesgue covering —, 6 combinatorial —, 50, 51 covering —, 5 large inductive —, 21 Lebesgue —, 6 mean projective —, 128 mean topological —, 111 Menger-Urysohn —, 21 small inductive —, 21 stable topological —, 110 topological —, 5 dimε (X, d), 78 Directed set, 186 Direction of an affine subspace, 49 Discrete topology, 6 Dispersion point, 102 Distal dynamical system, 155 Dynamical system, 192 distal —, 155 minimal —, 140 topologically mixing —, 136 topologically transitive —, 140 Dynamical systems topologically conjugate —, 112

E Embedding — of dynamical systems, 140, 192 topological —, 83 Entropy — of a dynamical system, 120, 220 — of a subshift, 137, 220 ε-dense subset, 146 ε-injective map, 76 Erdös space, 88 Even subshift, 136 Expansive homeomorphism, 137 Exponential growth group of —, 189 Extension — of a map, 71 — of groups, 164 Extremal point, 50 Extremally disconnected space, 48

F Face of a simplex, 50 Fan Knaster-Kuratowski —, 91 punctured Knaster-Kuratowski —, 91 Finer cover, 4

Index Finite type subshift of —, 129 Finitely generated group, 157 First-countable space, 35 Fixed point, 139 Flow, 153 Følner — condition, 161 — exhaustion, 188 — net, 187 — sequence, 158 Free — action, 221 Free group, 169 base of a —, 169

G General position, 80 Geometric realization of a simplicial complex, 65 Golden mean subshift, 136 Greatest lower bound, 95 Group — of exponential growth, 189 — of subexponential growth, 189 amenable —, 158, 161 Baumslag-Solitar —, 166 finitely generated —, 157 free —, 169 Heisenberg —, 166 Hopfian —, 187 linear —, 186 locally P —, 161 residually finite —, 46, 220 solvable —, 167 symmetric —, 162

H Heisenberg group, 166 Homeomorphism expaansive —, 137 Hopfian group, 187 Hyperplane affine —, 49

I Induced topology, 7 Invariant subset, 114, 191 Isolated point, 30 Iterates of a map, 110

Index J Join, 107

K Knaster-Kuratowski — fan, 91 — punctured fan, 91 k-simplex, 50

L Large inductive dimension, 21 Least upper bound, 95 Lebesgue — dimension, 6 — number, 10 Left — -coset, 197 — -translate, 157 Length of a chain of simplices, 54 Lindelöf space, 34 Linear group, 186 Locally connected space, 48 Lower bound, 95

M Map α-compatible —, 75 ε-injective —, 76 Mean, 186 — topological dimension, 111, 194 bi-invariant —, 186 left-invariant —, 186 right-invariant —, 186 Mesh, 10 — of a simplicial complex, 57 Middle third, 27 Minimal dynamical system, 140

N Neighborhood base, 31 Nerve, 73 Net, 186 Følner —, 187 Normal — form, 169 — space, 12

O Open

231 — cover, 4 — star, 53 Orbit, 139 Order — at a point of a family of subsets, 3 — of a family of subsets, 3 — topology, 95 — type, 96 Ordinal, 96 countable —, 97 infinite —, 97

P Partition, 4 — of unity, 72 Path-connected space, 66 Perfect space, 30 Perfectly normal space, 22 Periodic point, 139 Phase space, 192 Ping-pong principle, 172 Plank Tychonoff —, 97 Point almost-periodic —, 154 dispersion —, 102 extremal —, 50 fixed —, 139 isolated —, 30 periodic —, 139 Polyhedron, 52 Punctured Tychonoff plank, 98

Q Quasi-component, 41

R Regular space, 103 Relative amenability constant, 175 Relatively prime topology, 103 Residually finite group, 46, 220 Right-translate, 157

S Scattered space, 31 Second-countable space, 35 Semigroup, 119 Separable space, 36 Sequence subadditive —, 109

232 Sequentially-compact space, 104 Series derived —, 167 Set Fσ - —, 20 G δ - —, 20 well-ordered —, 96 Shift, 123, 204 — map, 123 Sierpinski triangle, 25 Simplex — in Rn , 50 — of an abstract simplicial complex, 64 barycenter of a —, 54 combinatorial dimension of a —, 50 face of a —, 50 open —, 51 Simplicial complex abstract —, 64 barycentric subdivision of a —, 57 geometric realization of a —, 65 mesh of a —, 57 Simplicial complex of Rn , 51 combinatorial dimension of a —, 51 support of a —, 51 vertex of a —, 51 Small inductive dimension, 21 Solvable group, 167 Sorgenfrey — line, 99 — plane, 100 Space — path-connected, 66 accessible —, 9 Bing —, 94 Cantor —, 28 completely normal —, 22 connected —, 9 Erdös —, 88 extremally disconnected —, 48 first-countable —, 35 Lindelöf —, 34 locally connected —, 48 normal —, 12 perfect —, 30 perfectly normal —, 22 phase —, 192 regular —, 103 scattered —, 31 second-countable —, 35 separable —, 36 sequentially-compact space, 104 σ -compact —, 34

Index T1 - —, 9 T4 - —, 12 totally disconnected —, 40 totally separated —, 41 ultrametric —, 47 Urysohn —, 102 Stable topological dimension, 110 Star open —, 53 Subadditive — map, 119 — sequence, 109 Subgroup derived —, 167 Subshift, 127, 204 — of block-type, 131, 211 — of finite type, 129 even —, 136 golden mean —, 136 Subspace affine —, 49 Successor, 97 Support — of a map, 72 — of a simplicial complex of Rn , 51 Symmetric — difference, 188 — group, 162 Syndetic, 154

T Ternary expansion, 29 Tits alternative, 186 Topological embedding, 83 Topological entropy, 120, 220 Topologically conjugate dynamical systems, 112, 192 Topologically mixing, 136 — action, 219 Topologically transitive dynamical system, 140 Topology — discrete, 6 cofinite —, 6 induced —, 7 order —, 95 relatively prime —, 103 trivial —, 9 Totally disconnected space, 40 Totally separated space, 41 Translation number, 119 Trivial topology, 9

Index Tychonoff plank, 97 punctured —, 98

U Ultrametric space, 47 Upper bound, 95 Upper-density, 147 Urysohn space, 102

233 V Vertex — of a simplicial complex of Rn , 51 — of an abstract simplicial complex, 64 Virtually P , 168

W Well-ordered set, 96